NEET PHY CH1 Module 3 - Baluni Classes

Charge and its properties

invention

1. Origin of friction electricity 2. The word charge was given by 3. Two kinds of charges 4. Algebraic sign to charge 5. Like charges repels and unlike attracts 6. Numerical value of force between two charges 7. Methods of charging 8. Amount of induced charge 9. e/m of electron 10. Charge and mass (indirectly) of electron and quanta of

charge 11. Concept of line of force

related scientist

1. In 600 B.C. (Thales) 2. Dr. William Gillbert 3. Du Fay 4. Benjamin Frankline 5. Dr. William Gillbert 6. Coulomb by (torsion balance experiment) 7. Thomos Brown 8. Faraday ice pail experiment 9. J.J. Thomson (1887) 10. R.A. millikan (by oil drop exp.) 11. M. Faraday

•• Order of size of atom and nucleus are 10-10 m (Å) and 10-15 m (Fermi) respectively. Order of density of nucleus = 1017 kg/m3.

•• Classical radius of electron 2.8 × 10-15 m.•• Millikan calculate quanta of charge by ‘Highest common

factor’ (HCF) method and it is = charge of electron.•• Smallest unit of charge is Franklin and largest is Faraday

1 C = 3 × 109 St. C, 1 Ab C = 10 C, • 1 Faraday = 96500 C.

•• Charge is invariant (it does not depends on speed).•• Attraction may take place between two likely charged

objects, generally it happens when difference in their charges is so large.

gold Leaf electroscope (gLe)

1. Any charged body (+ or -) come in closed to uncharged GLE, leaves diverges.

2. If a charged body come in close to charged GLE, leaves diverges if like object and converges if unlike object.

3. If X-ray incident on GLE, leaves always collapses, whatever be the charge on leaves.

•• Amount of induced charge ≤ inducing charge, equality holds only for metal.

•• If a body is charged either positive or negative, its vol-ume increases and density decreases.

•• Mass, length and time depends on state of rest or motion (Theory of relativity) while charge (also phase) not depends.

•• If quark particles are exist in nature, quantization of

charge still valid, but new quanta changes to e

3.

Coulomb Law and Factors affecting it

•• Electric force between two charges not depends on neighbouring charges.

•• If a dielectric slab (∈r) of thickness t is placed between two charges (distance d), force decreases.

F = Q Q

r1 2

024π ∈

where r = d - t + t ∈r

•• When two charges (Q1, Q2) are placed some distance apart. Neutral point is nearer to smaller charge and in between Q1 and Q2 if charges are like and away from charge if charges are unlike.

electrostatics

1Chapter

NEET_PHY CH1_Module 3.indd 1 3/9/2018 1:43:18 PM

1.2 Chapter 1

•• System of following charges, is in equilibrium, if:

(a) Q Q1 Qa a

QQ

1 4= −

(b)

Q Q

Q

Q2

a

a

a QQ

23

= −

•• For a three charge system it is not possible that all charges are in stable equilibrium.

•• Coulombs law is similar to newton’s gravitation law.

•• If Fg = Fe for two identical charges then q

mG= ∈4 0π

electric Field

•• For measuring �E practically a test charge (+ve) of mag-

nitude much less than the source charge should be used.•• Electric force on a charge in uniform E is constant and

hence acceleration is constant, so equations of motion

can be used (acceleration aqE

m= )

•• Electric field due to linear charge distribution: (a) Finite wire:

P

+++++++

RM

λ

α Ep = 2

2

k

R

λ αsin

(b) Infinite wire (α = 180°):

PR

++++++

λ

Ep = 2k

R

λ

(c) Charged arc:

OR

++++

++++

+++

α

λ

EP = 2

2

k

R

λ αsin

(d) Charged ring of radius R:

O x P

R

Q

at an axial point

EkQx

R xP =

+( );

/2 2 3 2 if

x R EkQ

x

x R Ekqx

R

P

P

>> =

<< =

or

or

2

2

•• As x increases: �E due to ring first ↑ then ↓ and at

xR

=2

it is maximum.

•• Force between two parallel wires is F (on unit length)

= 2 1 2k

d

λ λ

•• A small metal ball is suspended in a uniform �E. If X-ray

beam falls on it, the ball will be deflected in the direction of

�E.

•• If both �E and

�g (gravity) presents, then use resultant

acceleration.

(a) �FE and

�Fg same direction a g

qE

m= +

(b) �FE and

�Fg are opposite a g

qE

m= −

(c) �FE and

�Fg are perpendicular a2 = g2 +

qE

m

2

For example, Time period of simple pendulum

TL

a= 2π (a = net acceleration)

•• If identical charges are placed on each vertices of a regu-lar polygon, then

�E at centre = Zero.

•• Strength of �E ∝ [Separation between electric lines]-1

potential and potential difference

•• Potential is a relative parameter, while potential dif-ference is absolute i.e., potential depends on reference chosen. That’s why potential of same object [+ve, -ve, neutral] may have (+ve) or (-ve) or zero potential.

•• Potential of earth is assumed zero. It can be assumed anything. If it is 100 V (say) then potential of each point rises by 100 V but potential difference between two points remains same.

•• Remember the only one relation by which work done in moving a charge is calculated.Wexternal = q[Vf - Vi]; Vf = Potential of final point, Vi = Potential of initial point

•• Equipotential surface is always ⊥ to electric field i.e., component of

�E along equipotential surface is zero.

Charge distribution

Point charge, any sphere

Plate, sheet Wire, rod, cylinder

Equipotential surface

Spherical Plane Cylindrical

•• Gradient of any scalar is a vector, �E = -(Potential

gradient), �F = -(Energy gradient)


electrostatics 1.3

•• (a) Potential due to ring, VP = kQ

R x2 2+

(b) Non-conducting sphere, Vin = kQ R r

R

( )3

2

2 2

3

−

Also λ ⋅ 2πR = Q and at centre, VkQ

Rk0 2= = λ π( ),

VkQ

RV

kQ

rs = =, out

•• Two unlike unequal charges are placed some distance apart. The locus of points having zero potential lying on a circle nearer to smaller charge.

•• Potential of each point on a conducting surface is same whatever be its shape (i.e., it is an EPS). But both E and surface charge density (σ) at sharp points is more.

important derivation

Charged ring

Electric Field

Suppose a ring of radius a have positive charge q. This charge q is uniformly distributed over its circumference.

Linear charge density λπ

= =q

a2

Charge

Length

x

r

dE sin

dE sin2dE cos

dE

dE

dl

P

dlA B

a

+++++++++++++++++

+++++++++++++++++

θ

θθ

θ

θ

Let a very small element AB of length dl have charge,

dq = λdl = q

adl

2π⋅

Electric field intensity due to this charge (dq) at P,

dEdq

r=

4 02πε

Another element of length dl but dimetrically oppo-site also produce electric field intensity dE.

The vertical components will cancel each other and horizontal components will be added. So net electric field at point P due to whole ring (take small elements at differ-ent place on ring) can be calculated by adding all horizontal components.

So, resultant electric field due to one element = dE cos θ.

So, electric field due to whole ring,

E dE= Σ cos ,θ or Edq

r= ∑

4 02πε

θcos

[∵ dqq

adl=

2π and cosθ =

x

r]

Or Eq

adl

r

x

r

q

a

x

rdl= ⋅ ⋅ = ⋅∑ ∑

2

1

4 2 402

03π πε π πε

= = =+

q

a

x

ra

qx

rE

qx

a x2 42

4 403

03

02 2 3 2π πε

ππε πε

. .( )

∵ r a x= +

2 2

Electrical Potential

Let electric potential at point P due to small element dl is dV.

dVdq

r=

4 0πε

(Distance between small element and point P = r)Electric potential due to whole ring,

V dVdq

r rdq= = =∑∑ ∑

4

1

40 0πε πε

Vq

r=

4 0πε or V

q

a x=

+4 02 2 1 2πε ( )

motion of Charged particle in Uniform electric Field

l = Effective length of electric fieldL = Distance of screen from centre of electric field

parabolic path

(q, m, v)

vy = 0exit point

E

Y

ux = v

(0, 0) y

l

x

Particle at time t, for x direction, x = vt and for y direction,

y = uyt + 1

2 ayt

2

∵ uy = 0; t = x

v; ay =

qE

m


1.4 Chapter 1

∴ y = 0 + 1

2

2qE

m

x

v

or y = qE

mv2 2

x2 this is equation of parabola

Special results: 1. Time taken by the particle to cover length of electric

field, in other words, time for which particle moves

under the influence of electric field, T = l

v 2. Total deviation in the trajectory under electric field,

y = 1

2

2qE

m

l

v

3. For proton, duetron, and α-particle (When KE is same) y ∝ q for proton and deutron → track same 4. For electron and proton if KE same then y - same Curvature, sharpness of curve → same

mixing of identical Charged tiny drops

Let , number of tiny drops = N

R

N tiny drops Big drops

for each tiny drop for Big drop (r, q, σ, E, V) (R, Q, σB, EB, VB) 1. Charge conservation, Q = Nq

2. Volume conservation, 4

3π ⋅ N · r3 =

4

3π ⋅R3

Hence, R = N1/3r, Q = Nq

∴ σB = N1/3 σ, EB = N1/3 E, VB = N2/3 V•• When a soap bubble is charged (either +ve or -ve) then

the size is (radius) somewhat increases and +ve charger M ↓ ; -ve charger M ↑.

Electric Lines of Force (ELF)

Electric lines of electrostatic field have following properties: 1. Imaginary 2. Can never cross each other 3. Can never be closed loops 4. The number of lines originating or terminating on a

charge is proportional to the magnitude of charge. In rationalized MKS system (1/ε0) electric lines are asso-ciated with unit charge, so if a body encloses a charge q, total lines of force associated with it (called flux) will be q/ε0.

5. Total lines of force may be fractional as lines of force are imaginary.

+ BA –

qA > qB

+ –

6. Lines of force ends or starts normally at the surface of a conductor.

7. If there is no electric field there will be no lines of force.

8. Lines of force per unit area normal to the area at a point represents magnitude of intensity, crowded lines represent strong field while distant weak field.

9. Tangent to the line of force at a point in an electric field gives the direction of intensity. So a positive charge free to move follow the line of force.

Golden Key Points

•• Lines of force starts from (+ve) charge and ends on (-ve) charge.

•• Lines of force starts and ends normally on the surface of a conductor.

fixed point charge near

infinite metal plate

––––––

+

+ –+ –+ –+ –

+ –

E = 0E = 0

+ –

edge effect

σ σ

•• Lines of force do not exist inside a conductor (as field inside a conductor is zero) the field between the plates is as shown.

+ + ––+ + ––+ + ––+ + ––+ + ––

metal

Q −Q′ Q′ −Q+ + ––+ + ––+ + ––+ + ––+ + ––

dielectric

Q −Q′ Q′ −Q

E = 0 1Q′= Q(1 – )r∈

•• The lines of force never intersect each other due to superposition principle.

•• The property that electric lines of force contract longi-tudinally leads to explain attraction between opposite charges.

•• The property that electric lines of force exert lateral pressure on each other leads to explain repulsion between like charges.


electrostatics 1.5

electric Flux (f)

For open surface φ φ0 = = ⋅∫ ∫d E ds��

φ

φ π

c

c

E ds

qkq

= ⋅

=∈

∫��

� ( )

(

by defination of flux

or Gnetnet

0

4 aauss’s law)For closed surface

φ θc E S ES= ⋅ =��

cos

�S = Always normal to surface and pointed out wards.

Electric flux: 1. It is a real scalar quantity 2. Units: (V-m) and N-m2/C Dimensions: [ML3T -3A-1] 3. The value of f does not depend upon the distribution

of charges and the distance between them inside the closed surface.

The value of f is zero in the following circumstances: (a) If a dipole is (or many dipoles are) enclosed by a

closed surface. (b) Magnitude of (+ve) and (-ve) charges are equal

inside a closed surface. (c) If no charge is enclosed by the closed surface (d) Incoming flux (-ve) = Outgoing flux (+ve)

R

ds

Eds

E

dsE

fin = -πR2E and fout = πR2E, ftotal = 0

E

E

R

fin = fcircular = -πR2E and fout = fcurved

= πR2E, r ftotal = 0

E

ds

a

X

aa

Z

Y

ds

ds

fin = -a2E and fout = a2E

⇒ ftotal = 0

E

R

φ πin = –1

22R E and φ πout =

1

22R E. ftotal = 0

R q

E

Ekq

R=

2, φ π

π= ×

∈=

∈2

4 22

02

0

Rq

R

q

note

Here, electric field is radial.

q

fTotal = q

∈0

q

fhemisphere = q

2 0∈

q

R

fTotal = q

∈0

q

R

fcylinder = q

2 0∈

q fTotal = q

∈0


1.6 Chapter 1

q fcube = q

2 0∈

q fTotal = q

∈0

r

q

f = q

8 0∈

q fTotal = q

∈0

q

f = q

4 0∈

gauss theorem

The total flux linked with a closed surface is 1

0∈ times the

charge enclosed by the closed surface (Gaussian surface),

i.e., � ��

� E dsq⋅ =∈

∫0

This law is suitable for symmetrical charge distribu-tion and valid for all vector fields obeying inverse square law.

Gaussian Surface

1. Is imaginary surface 2. Is spherical for a point charge, conducting and non-

conducting spheres

3. Is cylindrical for infinite sheet of charge, conducting charge surfaces, infinite line of charges, charged cylin-drical conductors, etc.

We select a Gaussian surface such that determination

of E ds��

� ⋅∫ can be done in symplest way (symmetry)

Gauss Theorem•• Number of electric lines (ELF) related to 1 C (unit

charge) = 1/∈0.•• Electric lines contracts longitudinally and repels laterally.

Flux (f)

If enclosed charge (qen) is given then use, φ =∈qen

0

If external �E is given then use, f = E ds

�� ⋅∫

For an imaginary cube:

Positive of charge

Main centre

Face centre

Side centre Corner

Flux from cube

qε0

q2 0ε

q4 0ε

q8 0ε

•• Gauss theorem is applicable to all conservative field obeys inverse square law and can be derived from cou-lombs law.

�E calculated from it depends on charges

inside and outside the closed surface.

Charge distribution

Point charge, sphere

Wire, cylinder

Infinite plates

Gaussion surface

Spherical Cylindrical Cylindrical

••�E inside a conducting (solid, shell) sphere or cylinder is zero and outside, it is same as of non-conducting.

•• If temperature increases ∈r decreases. (as Eind ↓ and ∈r

= E0/(E0 - Eind); where E0 = main�E,

�Eind = induced

�E )

Shell Solid

Ein = 0

Es =kQ

R2

Eout =

Ein = Ein =

Es =

Eout =

Es =

Eout =kQr2

kQrR3 R2

kQr2

kQR2

Sphere

x = R

E

rx = R

E

r

Solid Hollow

2kRkr

Cylinder

x = R

E

rx = R

E

r

2k r

E due to nonconducting

Ein = 0

Es = R

Eout = r

Ein∝r λ

λ 2kλ

λ kλ


electrostatics 1.7

Electric Field Due to Solid Conducting or Hollow Sphere

For outside point (r > R): Using Gauss’s theorem,

� �

� E dsq

⋅ =∫Σε0

P

E

+

+

+

+

+

+

+ +

+

+

+

+

O

Gaussian surface

R

r

dA

∵ At every point on the Gaussian surface, � �E ds||

� �E ds⋅ = E ds cos 0° = E ds

∴ E dsq

⋅ =∫�Σε0

[E is constant over the gaussian surface]

E rq

× =4 2

0

πε

or Eq

rp =

4 02πε

For surface point (r = R):

Eq

RS =

4 02πε

For inside point (r < R): Because charge inside the con-ducting sphere or hollow is zero. (i.e., Σq = 0) So

E dsq��

� ⋅ = =∫Σε0

0 Ein = 0

electric Field due to solid non Conducting sphere

Outside (r > R): From Gauss’s theorem

P

E

O

Gaussian surface

R

r

ds

� �� E ds

q

s

⋅ =∫Σε0

E rq

× =4 2

0

πε

Eq

rP =

4 02πε

At surface (r = R):

Eq

RS =

4 02πε

Put r = R

Inside (r < R): From Gauss’s theorem,

� �� E ds

q

s

⋅ =∫Σε0

P

E

O

Gaussian surface

R

rds

Where Σq charge contained within Gaussian surface of radius r,

E rq

( )4 2

0

πε

=Σ

⇒ =Er

λπε2 0

(1)

As the sphere is uniformly charged, the volume charge den-sity (charge/volume) ρ is constant throughout the sphere,

ρπ

=q

R4

33

Charge enclosed in Gaussian surface, Σq r= ⋅ρ π4

33

= q

Rr

( )4 3

4

333

/ ππ⋅

or qqr

R∑ =

3

3 put this value in Eq. (1)

Eqr

Rin =

1

4 03πε

Electric Field Due to an Infinite Line Distribution of Charge

Let a wire of infinite length is uniformly charged having a constant linear charge density λ.


1.8 Chapter 1

P is the point where electric field is to be calculated.Let us draw a coaxial Gaussian cylindrical surfaces

of length l.From Gauss’s theorem,

� � � � � �E dS E dS E dS

q

s s s

⋅ + ⋅ + ⋅ =∈∫ ∫ ∫1 2 3

0

� �E dS⊥ 1 so

� �E dS⋅ =1 0 and

� �E dS⊥ 2 so

� �E dS⋅ =2 0

E

l

E

E

ds1

ds2

ds3

Gaussiansurface

λ

+++++++++++++

E rlq

× =20

πε

[∵� �E dS||

3 ]

Charge enclosed in the Gaussian surface, q = λl.

so, E rll

× =20

πλε

⇒ =Er

λπε2 0

⇒ =Ek

r

2 λ where k =

1

4 0πε

electric potential due to hollow or Conducting sphere

At Outside Sphere

According to definition of electric potential, electric poten-tial at point P

R

r > R

P

V E dq

rdr E

q

r

r r

= − ⋅ − = =

∞ ∞

∫ ∫��

∵4 40

20

2πε πεout

Vq

rdr

q

r

q

r

r r

= − =

=

∞ ∞∫4

1

4

1

402

0 0πε πε πε

At Surface

V E drq

rdr E

q

r

R R

= − ⋅ = − =

∞ ∞

∫ ∫��

∵4 40

20

2πε πεout

Vq

rdr

q

r

R R

= −

=

∞ ∞∫4

1

4

1

02

0πε πε

Vq

R=

4 0πε

Inside the Surface

∵ Inside the surface E = 0, dV

dr= 0 or

V = constant ∵ EdV

dr= −

So, V = q

R4 0πε

electric potential due to solid non Conducting sphere

At outside sphere: Same as conducting sphere.At surface: Same as conducting sphere.Inside the sphere:

V E drr

= − ⋅∞∫��

or V E dr E drR

rR

= − +

∫∫∞

1 2

= −

+

∫∫∞

kq

rdr

kqr

Rdr

R

rR

2 3

V kqr

kq

R

rR

R

r

= − −

+

∞

1

23

2

V kqR

r

R

R

R= − − + −

1

2 2

2

3

2

3

Vkq

RR r= −

23

32 2[ ]

electric dipole

•• For a dipole, potential is zero at equator, while at any finite point E ≠ 0

•• In a uniform �E , dipole may feels a torque but not a force.

•• If a dipole placed in a field generated by a point charge, then torque on dipole may be zero.


electrostatics 1.9

DistributionPoint

charge Dipole Quadrapole Octapole

Potential proportional to r -1 r -2 r -3 r -4

E proportional to r -2 r -3 r -4 r -5

Force between Point

chargeDipole and

point charge Dipole-dipole

Proportional to r -2 r -3 r -4

Work Done in Rotating an Electric Dipole in an Electric Field

Suppose at any instant, the dipole makes an angle θ with the electric fi eld. The torque acting on dipole, d = 2l sin θτ = qEd = (q 2l sin θ)E = pE sin θThe work done in rotating dipole from θ1 to θ2

W d pE d= =∫ ∫τ θ θ θθ

θ

θ

θ

1

2

1

2

sin

–qE

qE

E–q

+q

A C

d2l

θ

B

W = pE (cos θ1 - cos θ2) = U2 - U (1)

(∵ U = -pE cos θ)

Potential Energy of an Electric Dipole in an Electric Field

The potential energy of a dipole is defi ned as work done in rotating a dipole from a direction perpendicular to the fi eld to the given direction.

W d pE d= =∫ ∫τ θ θ θθ θ

90

2

90

2

� �

sin [∵ τ = pE sin θ]

W = -pE cos θ

This work is stored in potential energy.

So, U W p E= = − ⋅� �

Electric Field Due to an Electric Dipole

At a point on the axis of a dipole:

Electric fi eld at P due to +q charge, E1 = kq

r l( )− 2

–q q

r

O

E1E2

P

ll

Electric fi eld at P due to -q charge, E2 = kq

r l( )+ 2

Net electric fi eld at P, E = E1 - E2

E = kq

r l

kq

r l( ) ( )−−

+2 2

E = kq rl

r l

×

−

42 2 2( )

[∵ p = q × 2l = Dipole moment]

E = 22 2 2

kpr

r l( )−

If r >>>l E = 2

3

kp

r

At a point on equatorial line of dipole:

Electric fi eld due to +q charge, Ekq

x1 2=

E P

x

–q qO

θ

θθ

θ

x

θE

2sin

θE1sin

E2

E1

r

l l

Electric fi eld due to -q charge, Ekq

x2 2=

Vertical component of E1 and E2 will cancel each other and horizontal components will be added.

So net electric fi eld at P,

E = E1cos θ + E2cos θ [∵ E1 = E2]

E = 2E1 cos θ = 2

2

kq

xcosθ

[∵ cosθ =l

x and x r l= +2 2 ]


1.10 Chapter 1

Ekql

x

kql

r l= =

+

2 23 2 2 3 2( )

or Ekp

r l=

+( )2 2 3 2

If r > > > l then Ekp

r=

3

Electric Potential Due to Dipole

At axial point:

Electric potential due to +q charge, Vkq

r l1 = −( )

– q q

r

OP

ll

Electric potential due to -q charge, Vkq

r l2 =−+( )

Net electric potential, V = V1 + V2 = kq

r l

kq

r l( ) ( )−+

−+

=×

−=

−

kq l

r l

kp

r l

22 2 2 2( )

If r > > > l then Vkp

r=

2

At equatorial point:

Electric potential of +q charge, V1 = kq

x

Electric potential of -q charge, V2 = −kq

x

P

xr

–q qO

x

l l

Net potential, V = v1 + v2 = kq

x

kq

x− = 0

So, V = 0

soLved eXampLes

eXampLe 1

Two conducting shells are placed co-centrically and a total charge Q is given to them so that their surface charge densi-ties are equal. Deduce an expression for potential at com-mon centre.

b

a

A

B

soLUtion

Let surface charge density of each is σ then charge of A + charge of B = Q

⇒ σ4πa2 + σ4πb2 = Q or σ = Q

a b4 2 2π ( )+ (1)

Now potential at centre = Potential due to A + Potential due to B

V0 = σ∈0

a + σ∈0

b = σ∈0

(a + b)

Now, put value of σ from eq. (1)

V0 = ( )

( )

a b Q

a b

+

∈ +4 02 2π

eXampLe 2

A point charge Q is placed at a point A(0, 0). Calculate electric fi eld at B(x, y) in vector form.

soLUtion

Electric fi eld is,

E = kQ

r2 in vector form � �E

kQ

rr

kQr

r=

=

2 3ˆ

∵�

rr

r=

r

B(x, y)

Y

XAO


electrostatics 1.11

Now �r = �

rB - �

rA = xi + yj and r2 = x2 + y2

So, �E =

kQ xi yj

x y

( )

( ) /

+

+2 2 3 2

eXampLe 3

Linear charge density of arc is λ and total angle at center α. Derive an expression for potential.

r

α

λ

O

soLUtion

Consider a small element of length dl and charge dq then dq = λ · dl Now potential at centre due to element is

r

O

dqdl

λ

dV = kdq

r

k dl

r

k

rdl= =

( )λ λ

To calculate total potential, integrate it

V dVk

rdl

k L

r= = =∫ ∫

λ λ

L = Total length of arc = rα. So, V = k

r

λ(r · α) therefore

V = kλα.

eXampLe 4

If separation between two point charges decreases then electric potential energy of system increases, whatever the sign of charges. Yes or No?

soLUtion

No [∵ It depends on sign of charges.]

eXampLe 5

Consider two spherical shells whose centre coincides each other figure. Potential at common centre is given by:

b

a

σσ

O

soLUtion

σ∈0

(a + b)

eXampLe 6

Linear charge density of arc is λ and total angle at centre is α. Calculate electric field at centre.

soLUtion

Consider two small elements as in Figure. Resultant E due to two elements is,

dET = 2dE cos θ = 2kdQ

r2

cos θ

r

OdE

2dEcos

dE

dq

dq

λ

++++++

++++

++++

++++

θ θθ

∵dEkdQ

r=

2

dET = 2

2

K dl

r

( )λcos θ (dQ = λdl)

rd

λ

θθ

++++++

+++

++

++++

+++

dl

= 2

2

K rd

r

λ θ( ) cos θ =

2K

r

λ (cos θ) dθ ∵ d

dl

rθ =

Now total electric field is,

ET = 2

0

2k

r

λ

θ

θ α

=

=

∫ cos θ dθ = 2

2

k

r

λ αsin

eXampLe 7

A region consists uniform electric field E and uniform gravity g perpendicular to each other. A particle of charge Q and mass m is starts to move in this region. Deduce an equation of trajectory.


1.12 Chapter 1

soLUtion

If E is in (positive x-direction)

Then ax = QE

m; x =

1

2

QE

mt2 (1)

and ay = g; y = 1

22gt (2)

Solving Eqs. (1) and (2), we can get equation of trajectory

y = mg

qEx

(Straight line)

eXampLe 8

A dipole is placed in uniform electric field �E at an angle θ.

For what value of θ: (A) Torque on dipole is maximum (B) Equilibrium is stable

soLUtion

(A) 90° (B) 0°

eXampLe 9

An infinite number of point charges each equal to q are placed at x = 1, 2, 4, 8, …. Calculate the electrostatic poten-tial at origin.

soLUtion

Electrostatic potential at a point distance r due to a charge

q is 1

4 0π ∈q

r

The net electrostatic potential at origin (x = 0) due to the infinite array is,

V = q

4

1

1

1

2

1

4

1

80π ∈+ + + +

�

= q

4

1

2

1

2

1

2

1

200 1 2 3π ∈+ + + +

�

= q

4 0π ∈⋅

1

1 12−( ) =

2

4 0

q

π ∈=

q

2 0π ∈

note

Sum of infinite series of GP, S = a

r( )1−

eXampLe 10

If numerical value of potential and field of a short dipole at a point r distance away from centre of dipole is equal then find out minimum value of r.

soLUtion

Given that V = | |�E

or kp

r

cosθ2 =

kp

r321 3+ cos θ

or cos θ = 1

1 3 2

r+ cos θ

or cos2θ = 12r

(1 + 3cos2θ)

or r2 cos2θ = 1 + 3cos2θ

or r2 = 1

2cos θ + 3

or r2 = sec2θ + 3Now minimum value of sec θ is 1 so r2

min = 1 + 3 = 4or rmin = 2 r

eXampLe 11

If electric potential energy of given system is positive then prove that 2Q > 3q.

soLUtion

U (system) = Sum of potential energy of all pairs

= k Q Q

a

( ) ( )2 -

k Qq

a

2 -

kQq

a =

kQ

a(2Q - 3q)

Q

a a

a(2Q) (–q)

Given U is positive means U > 0

⇒ kQ

a(2Q - 3q) > 0 ⇒ 2Q > 3q

eXampLe 12

Two point charges (+Q) and (-Q) are placed at (0, 2) m and (0, -2) m respectively. Show that the direction of electric field on X-axis is along (-y) direction.

(x, 0)

EEE–Q(0, –2)

Q(0, 2)


electrostatics 1.13

soLUtion

Direction of E due to +Q and -Q is as shown in figure. as both are equal in magnitude so resultant is in between of E+ and E. Means along (-y) direction.

eXampLe 13

Why two equipotential surfaces cannot intersect?

soLUtion

An equipotential surface is normal to the electric field intensity. If two equipotential surfaces intersect, then at the point of intersection, there will be two directions of electric field intensity which is not possible.

eXampLe 14

Write down one difference and two similarity between electric force and gravitational force.

soLUtion

Similarity: (Any two) 1. Both are conservative in nature. (i.e., for both work

done in a loop is zero.) 2. Both are central forces. 3. Both follows inverse square law. 4. Both are infinite range force. 5. Both follows newton’s third law.

Differences: (Any one) 1. Gravitational force is very weak as compare to electric

force. 2. Gravitational force is always attractive while electric

force may be attractive or repulsive 3. Electric force depends on medium, while gravitational

force does not depend on medium. 4. Gravitational force is an interaction between mass

particles while electric force is interaction between charges.

eXampLe 15

Is it possible that flux from an imaginary closed surface is zero even when electric field on this surface is non-zero. If yes then give one example.

soLUtion

Yes (Example: A dipole is placed in closed surface)

eXampLe 16

A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at distance 3r from the centre is V. Find out the electric field intensity at a distance 3r from the centre.

soLUtion

Given V1 - V2 = V

kQ

r

kQ

rV− =

3

3r

r 1 2

2

3

kQ

r = V (kQ) =

3

2

Vr (1)

Now required electric field, E =kQ

r

Vr

r( )

( )

( )3

3 2

32 2=

/ =

3

2 9 2

Vr

r×

E = V

r6

eXampLe 17

A ring of radius R is charged uniformly by λ (C/m). Derive a relation between potential on axis with distance x from centre and hence obtain potential at centre.

soLUtion

Consider a small element of length dl then

dQ = λdl

Now potential due to small element dQ = (as point charge)

x

r

P

dQ

R

Q

++++++++++++++++++++++++

+++++++++++++++++++++

dV = kdQ

r (Distance is r)

or dV = k

r(λdl)

Potential is scalar, and distance of all elements is same (r), so


1.14 Chapter 1

V dVk dl

r

k

rdl

k

r= = = =∫ ∫ ∫

( )λ λ λ(2πR)

=∈

=∈ +

λ λR

r

R

R x2 20 02 2

Now for centre, x = 0 ∴V0 = λ

2 0∈

eXampLe 18

Plot the following graphs:

(A) Electric field inside a conducting sphere with distance from centre.

(B) E versus 1

r where E is electric field due to a point

charge and r is the distance from charge. (C) Electric potential energy (U) of a pair of 2 like charges

with distance (r) between charges.

soLUtion

(A)

r = 0 r = Rr

E

(B) E

1r

(C) U

r

eXampLe 19

In a region �E = 2 i + 3 j + 4 k then potential difference

between A(0, 0, 0) and B(1, 2, 1) will be:

soLUtion

Vf - Vi = - � �E dr⋅ , dr OB OA

� � �= −

VB - VA = -(2 i + 3 j + 4k ) ⋅ (i + 2 j + k )

= -(2 + 6 + 4) = -12 VSo potential difference between A and B is

= VA - VB = 12 V

eXampLe 20

For the given figure. charge q1 = 2 × 10-8 C and q2 = -0.4 × 10-8 C. A charge 7q3 of 0.2 × 10-8 C is taken along arc of circle from C to D, then potential energy of charge q3 is decreased by:

U

ARC

q3

q1 q2

80 cm 100 cm

60 cm D

BD = 20 cmC

AB

soLUtion

EPE of q3 at C = q3 k q k q1 2

0 8 1.+

EPE of q3 at D = q3

k q k q1 2

0 8 0 2. .+

or EPE at C and D are 3.78 × 10-7 J and 0.9 × 10-7 J.

percentage of decrement = 2 88

3 78

.

. × 100 = 76% (decreases,

76%)

eXampLe 21

Three charges -q, +q and +q are situated in XY-plane at points (0, -a), (0, 0) and (0, a) respectively. The potential at a point distant r (r > a) in a direction making an angle θ from Y-axis will be:

soLUtion

Potential at P due to charge q (placed at origin) is,

V0 = kq

r

q(0, a)

(0, –a) –q

OX

dipole momentP = q(2a)

Y

θ

Now potential due to remaining charge system (dipole) can be written as,


electrostatics 1.15

Vd = kP

r2 cos(θ) = kq a

r

⋅22

cosθ

Hence, VP = V0 + Vd = kq

r1

2+

a

r

cosθ

eXampLe 22

Two similar balloons filled with helium gas are tied to two 1 m long strings. A body of mass m is tied to another ends of the strings. The balloons float in air at distance r. If the amount of charge on the balloons is same then the magni-tude of charge on each balloons will be:

soLUtion

For mass m, 2T cos θ = mg

for balloon, T sin θ = Kq

r

2

2

T T

Tθ

θ

θ

θ

TsinFE

mg

2 cot θ = mgr

kq

2

2 or q =

mgr

k

2 1 2

2tan

/

θ

eXampLe 23

What is the dipole moment of the system shown in figure?

aa

a –q

–q

2q

A

soLUtion

There are two diploes of �P = q (a)

So, Pnet = 3 p = 3 qa.

eXampLe 24

A proton and a α-particle are situated at r distance apart. At very large dis-tance apart (when released.) the KE of α-particle will be:

soLUtion

Initially total energy of system, Ei = ke e

r

ke

r

⋅=

2 2 2

2e

4 mem

Pα

Momentum of both particle at very large distance is same.

Hence, E α 1

m

So divide total energy in inverse ratio of mass, Ep =

4

5Ei =

4

5

2 2ke

r

, Eα = 1

5Ei =

1

5

2 2ke

r

.

eXampLe 25

Graph shows variation of potential vs distance. Calculate electric field at various points.

E = −dv

dr = -Slope = -tan θ

soLUtion

1 2 3 4 5 6 r

1

2

V(volts)

A BE=0

E=–veE=+ve

C0

Distance (m) E (V/m)

r = 1 -1

r = 1.5 -1

r = 2.5 0

r = 5 1

eXampLe 26

For systems to be in equilibrium Q = ?

a a

a

a

Q

q q

q q

aa

aq q

q

Q

q

Q

qa a

60°

P

P

A


1.16 Chapter 1

soLUtion

For square, Q = − +( )2 2 1

4q,

For triangle, Q = −q

3

For linear, Q = −q

4

eXampLe 27

Calculate protonic charge in 100 cc of water.

soLUtion

1 cc = 1 g for water as density = 1 g/cm3

Now number of atoms in 18 g (atm weight) = 6.023 × 1023 and each molecule of H2O contains 10 proton (8 of oxygen + 2 of H2).

So number of proton in 100 g water = 6 023 10

18100

23. ××

× 10 = 3.3 × 1025

Hence, protonic charge = 3.3 × 1025 × 1.6 × 10-19 = 5.4 × 106 cb.

eXampLe 28

Find VA, VB, VC, and If VA = VC then what is the required condition?

a

bc

A

B

σ

σ

−σ C

soLUtion

Va b c

a b cA = ∈−∈

+∈

=∈

− +σ σ σ σ

0 0 0 0

( )

Va

b

b cB =

∈−∈

+∈

σ σ σ2

0 0 0

;

Va

c

b

c

c a b c

cC =∈

−∈

+∈

=∈

− +

σ σ σ σ2

0

2

0 0 0

2 2 2

⇒ Now if, VA = VC ⇒ (a - b + c)

= a b c

c

2 2 2− +⇒ c (a - b) = a2 - b2 ⇒ c = a + b

eXampLe 29

Compare field, potential and surface charge density at A, B, C.

soLUtion

Surface of a metal is an equipotential surface (EPS) so charge on irregular shaped metal distributes to create same potential on surface. ⇒ VA = VB = VC Now radius of curvature (R) for straight line = ∞

RC > RB > RA Now Q

R

Q

R

Q

RA

A

B

B

C

C

= = or (Q ∝ R)

Electric Field, E ∝Q

R2 or E ∝1

R⇒ EA > EB > EC

METAL

+++

++ + + + +

++

+++ +

++++++++

+++++++++++++++++++++++

+++

AB

C

Surface charge density, σ ∝ Q

R2

⇒ σA > σB > σC

So, if there is any possibility of charge leakage, it starts from point A. (corona discharge)

note

Remember that E and σ at sharp points is more while potential is same.

eXampLe 30

Two (-ve) charge, each of magnitude q are situated at 2r distance apart. A (+ve) charge q is lying at the centre between them. The potential energy of the system is U1. If the two nearest charges are mutually interchanged and the

potential energy becomes U2, then U

U1

2

will be:

soLUtion

Ukq

r

kq

r

kq

r

kq

r1

2 2 2 2

2

3

2=−

+−

+ =−

–q

r r

–qq


electrostatics 1.17

Ukq

r

kq

r

kq

r

kq

r2

2 2 2 2

2 2=−

+ − =−

so U

U1

2

3=

q

r r

–q–q

eXampLe 31

If σ = -2 × 10-6 C/m2 Calculate d such that electron strikes the plate with zero velocity.

soLUtion

Kinetic energy of e- is consumed in work done against elec-tric force (EK = KE of e-)

M

E

T

A

L

––––––––––––––

plate ––

100 eV

d

e–

σ

electron gun

Ek = Force × displacement = (eE)d

so Ek = (e) E ⋅ d. = e ⋅ σ∈0

d

⇒ d = E

ek ∈0

σ.

LeveL i

objective problems—neet

(C) 1.6 × 10-19 2

11

2

− Coulomb

(D) 1 6 10

11

2

19

2

. ×

−

− Coulomb

5. A stationary electric charge produces: (A) Only electric fi elds (B) Only magnetic fi eld (C) Both electric as magnetic fi eld (D) Neither electric nor magnetic fi eld

6. Which one of the following statement regarding elec-trostatics is wrong?

(A) Charge is quantized. (B) Charge is conserved. (C) There is an electric fi eld near an isolated charge at rest. (D) A stationary charge produces both electric and

magnetic fi elds.

7. An accelerated or decelerated charge produces: (A) Electric fi eld only (B) Magnetic fi eld only (C) Localized electric and magnetic fi elds (D) Electric and magnetic fi elds that are radiated

Section A: Coulomb’s law, Electrostatic Equillibrium + Properties of Charge

1. The unit of charge is coulomb in SI system and esu of charge (or stat coul) in CGS system 1 coulomb equals:

(A) (3 × 109) esu (B) (1/3 × 109) esu (C) (1/3 × 108) esu (D) (9 × 109) esu

2. The relative strengths of gravitational, electromagnetic and strong nuclear forces are:

(A) 1 : 1039 : 1036 (B) 1 : 1036 : 1039

(C) 1 : 10-26 : 10-39 (D) 1 : 10-39 : 10-36

3. Fg and Fe represent the gravitational and electrostatic force respectively between two electrons situated at some distance. The ratio of Fg to Fe is of the order of:

(A) 1036 (B) 101

(C) 10° (D) 10-43

4. An electron at rest has a charge of 1.6 × 10-19 C. It starts moving with a velocity v = c/2, where c is the speed of light, then the new charge on it is:

(A) 1.6 × 10-19 Coulomb

(B) 1.6 × 10-19 11

2

2

−

Coulo mb

eXerCises


1.18 Chapter 1

8. If a glass rod is rubbed with silk, it acquires a positive charge because:

(A) Protons are added to it (B) Protons are removed from it (C) Electrons are added to it (D) Electrons are removed from it

9. One quantum of charge should be at least being equal to the charge in coulomb:

(A) 1.6 × 10-17 (B) 1.6 × 10-19 (C) 1.6 × 10-10 (D) 4.8 × 10-10

10. When the distance between two charged particle is halved, the force between them becomes:

(A) One fourth (B) One half (C) Double (D) Four times

11. The force between an α -particle and an electron sepa-rated by a distance of 1 Å is:

(A) 2.3 × 10-8 N attractive (B) 2.3 × 10-8 N repulsive (C) 4.6 × 10-8 N attractive (D) 4.6 × 10-8 repulsive

12. Two similar and equal charges repel each other with force of 1.6 N, when placed 3 m apart. Strength of each charge is:

(A) 40 µC (B) 20 µC (C) 4 µC (D) 2 µC

13. There are two charges +1 micro-coulomb and +5 micro-coulomb, the ratio of force on them will be:

(A) 1043 (B) 1 : 1 (C) 10° (D) 10-43

14. The force between two point charges placed in vacuum at distance 1 mm is 18 N. If a glass plate of thickness 1 mm and dielectric constant 6, be kept between the charges then new force between them would be:

(A) 18 N (B) 108 N (C) 3 N (D) 3 × 10-6 N

15. Two charges are at distance (d) apart in air. Coulomb force between them is F. If a dielectric material of dielectric constant (K) is placed between them, the Coulomb force now becomes:

(A) F/K (B) FK (C) F/K2 (D) K2F

16. A certain charge Q is divided at first into two parts, (q) and (Q - q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then:

(A) (Q/q) = (4/1) (B) (Q/q) = (2/1) (C) (Q/q) = (3/1) (D) (Q/q) = (5/1)

17. Two point charges in air at a distance of 20 cm from each other interact with a certain force. At what dis-tance from each other should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction:

(A) 8.94 × 10-2 m (B) 0.894 × 10-2 m (C) 89.4 × 10-2 m (D) 8.94 × 102 m

18. The permittivity ∈0 of vacuum is 8.86 × 10-12 C2/N-m2 and the dielectric constant of water is 81. The permit-tivity of water in C2/N-m2 is:

(A) 81 × 8.86 × 10-12

(B) 8.86 × 10-12

(C) (8.86 × 10-12)/81 (D) 81/(8.86 × 10-12)

19. The force between two point charges in vacuum is 15 N, if a brass plate is introduced between the two charges, then force between them will:

(A) Becomes zero (B) Remains the same (C) Becomes 30 N (D) Becomes 60 N

20. A charge Q is divided in two parts Q1 and Q2 and these charges are placed at distance R there will be maxi-mum repulsion between them, when:

(A) Q2 = (Q/R), Q1 = Q -(Q/R)

(B) Q2 = (Q/3), Q1 = (2Q/3)

(C) Q2 = (Q/4), Q1 = (3Q/4)

(D) Q1 = Q2 = Q/2

21. A mass particle (mass = m and charge = q) is placed between two point charges of charge q separation between these two charge is 2L. The frequency of oscillation of mass particle, if it is displaced for a small distance along the line joining the charges:

(A) q

m L2

1

03π πε

(B) q

m L2

4

03π πε

(C) q

m L2

1

4 03π πε

(D) q

mL2

1

16 03π πε

22. ABC is a right angle triangle AB = 3 cm, BC = 4 cm, charges +15, +12, -12 esu are placed at A, B and C respectively. The magnitude of the force experienced by the charge at B in dyne is:

(A) 125 (B) 35 (C) 22 (D) 0

23. Equal charges of each 2 µC are placed at a point x = 0, 2, 4, and 8 cm on the X-axis. The force experienced by the charge at x = 2 cm is equal to:

(A) 5 N (B) 10 N (C) 0 N (D) 15 N


electrostatics 1.19

24. Three equal charges (q) are placed at corners of a equi-lateral triangle. The force on any charge is:

(A) Zero (B) 32

2

Kq

a

(C) Kq

a

2

23 (D) 3 3

2

2

Kq

a

25. Five point charges, each of value +q coulomb, are placed on five vertices of a regular hexagon of side L m. The magnitude of the force on a point charge of value -q coulomb placed at the centre of the hexagon is:

(A) kq

L

2

2 (B) 5 kq

L

2

2

(C) 32

2

kq

L (D) Zero

26. Two small balls having equal positive charge Q (Coulomb) on each are suspended by two insulating strings of equal length L m, from a hook fixed to a stand. The whole set up is taken in a satellite in to space where there is no gravity (state of weightless-ness) Then the angle (θ) between the two strings is:

(A) 0° (B) 90° (C) 180° (D) 0° < θ < 180°

27. Two similar charge of +Q, as shown in figure are placed at A and B. -q charge is placed at point C mid-way between A and B. -q charge will oscillate if:

C

–q

+Q

A

+Q

B

(A) It is moved towards A (B) It is moved towards B (C) It is moved upwards AB (D) Distance between A and B is reduced

28. Two equal negative charge (-q) are fixed at the points (0, a) and (0, -a) on the Y-axis. A positive charge (Q) is released from rest at the point (2a, 0) on the X-axis. The charge Q will:

(A) Execute simple harmonic motion about the origin (B) Move to the origin and remains at rest (C) Move to infinity (D) Execute oscillatory but not simple harmonic motion

29. Four charges are arranged at the corners of a square ABCD, as shown. The force on +ve charge kept at the centre of the square is:

+q –q

+2q–2qA

B C

D

(A) Zero (B) Along diagonal AC (C) Along diagonal BD (D) Perpendicular to the side AB

30. Two free positive charges 4q and q are a distance l apart. What charge Q is needed to achieve equilibrium for the entire system and where should it be placed form charge q?

(A) Q q=4

9 3(negative) at

l

(B) Q q=4

9 3( )positive at

l

(C) Q = q (positive) at l3

(D) Q = q (negative) at l3

31. Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be:

–q 3q

–2q–2q

q 2q

A B

DE

F

(A) Zero (B) Along OF (C) Along OC (D) None of these

Section B: Electric Field, Questions on Constant Electric Field, Electric Field due to Uniformly Charged Rod, Time Period of SHM + Electrostatics, Electric Field Due to Uniformly Charged Ring, Disc, Sheet

32. If Q = 2 C and force on it is F = 100 N, Then the value of field intensity will be:

(A) 100 N/C (B) 50 N/C (C) 200 N/C (D) 10 N/C


1.20 Chapter 1

33. Which one of the following relations is correct? (A) 1 N/C = 108 V/m (B) 1 N/C = 10-6 V/m (C) 1 N/C = 1 V/m (D) 1 N/C = 10-8 V/m

34. Four equal but like charge are placed at four corners of a square. The electric field intensity at the center of the square due to any one charge is E, then the resultant electric field intensity at centre of square will be:

(A) Zero (B) 4E (C) E (D) 1/2E

35. If mass of the electron = 9.1 × 10-31 kg. Charge on the electron = 1.6 × 10-19 C and g = 9.8 m/s2. Then the intensity of the electric field required to balance the weight of an electron is:

(A) 5.6 × 10-9 N/C (B) 5.6 × 10-11 N/C (C) 5.6 × 10-8 N/C (D) 5.6 × 10-7 N/C

36. A point charge 50 µC is located in the XY-plane at the point of position vector

�r i j0 2 3= +ˆ ˆ. What is the elec-

tric field at the point of position vector �r i j= −8 5ˆ ˆ ?

(A) 1200 V/m (B) 0.04 V/m (C) 900 V/m (D) 4500 V/m

37. A point charge q is placed at origin. Let � �E EA B, and �

EC be the electric field at three points A(1, 2, 3), B(1,

1, -1) and C(2, 2, 2) due to charge q. Then:

[i] � �E EA B⊥ [ii] | | | |

� �E EB C= 4

Select the correct alternative: (A) Only [i] is correct (B) Only [ii] is correct (C) Both [i] and [ii] are correct (D) Both [i] and [ii] are wrong

38. Two charges 4q and q are placed 30 cm. apart. At what point the value of electric field will be zero?

(A) 10 cm away from q and between the charges. (B) 20 cm away from q and between the charges. (C) 10 cm away from q and outside the line joining the

charge. (D) 10 cm away from 4q and outside the line joining

them.

39. A pendulum bob of mass 80 mg and carrying a charge of 2 × 10-8 C is at rest in a horizontal uniform electric field of 20,000 V/m. Find the tension in the thread of pendulum.

(A) 8.8 × 10-2 N (B) 8.8 × 10-3 N (C) 8.8 × 10-4 N (D) 8.8 × 10-5 N

40. Six charges +Q each are placed at the corners of a reg-ular hexagon of side (a), the electric field at the centre of hexagon is:

(A) Zero (B) 1

4 0π ∈⋅6 2

2

Q

a

(C) 1

4 0π ∈⋅Q

a

2

2 (D)

14 0π ∈

⋅6 2

2

Q

a

41. Four charges +q, +q, -q and -q are placed respectively at the corners A, B, C and D of a square of side (a), arranged in the given order. Calculate the intensity at (O) the centre of the square.

(A) 4

4 20

2πε ⋅a

q (B)

4 2

4 02

q

aπε ⋅

(C) πε0

2

4 2

⋅a

q (D)

4 2

02

q

aπε ⋅

42. Two charged spheres A and B are charged with the charges of +10 and +20 C respectively and separated by a distance of 80 cm. The electric field at a point on the line joining the centres of the two spheres will be zero at a distance from sphere A:

(A) 20 cm (B) 33 cm (C) 55 cm (D) 60 cm

43. Two charges 9e and 3e are placed at a distance r. The distance of the point where the electric field intensity will be zero is:

(A) r

1 3+

from 9e charge

(B) r

1 1 3+

/ from 9e charge

(C) r

1 3−

from 3e charge

(D) r

1 1 3+

/ from 3e charge

44. A proton is first placed at A and then at B between the two plates of a parallel plate capacitor charged to a PD of V volt as shown. Then force on proton at A is:

–+++++++++

–––––

––

A

B


electrostatics 1.21

(A) More than at B

(B) Less than at B

(C) Equal to that at B

(D) Nothing can be said

45. A ring of radius (R) carries a uniformly distributed charge +Q. A point charge -q is placed on the axis of the ring at a distance 2R from the centre of the ring and released from rest. The particle:

(A) Becomes in rest condition immediately

(B) Executes simple harmonic motion

(C) Motion is not SHM

(D) Come at the centre of ring immediately

46. A small circular ring has a uniform charge distribution. On a far-off axial point distance x from the centre of the ring, the electric field is proportional to:

(A) x-1 (B) x-3/2

(C) x-2 (D) x5/4

47. A non-conducting ring of radius R has uniformly dis-tributed positive charge Q. A small part of the ring, of length d, is removed (d<<R). The electric field at the centre of the ring will now be:

(A) Directed towards the gap, inversely proportional to R3

(B) Directed towards the gap, inversely proportional to R2

(C) Directed away from the gap, inversely propor-tional to R3

(D) Directed away from the gap, inversely propor-tional to R2

48. Which of the following statements concerning the electrostatics is correct?

(A) Electric line of force never intersect each other. (B) Electric lines of force start from positive charge

and end at the negative charge. (C) Electric lines of force start or ends perpendicular

to the surface of a charged metal. (D) All of these

49. The tangent drawn at a point on a line of electric force shows the:

(A) Intensity of gravity field (B) Intensity of magnetic field (C) Intensity of electric field (D) Direction of electric field

50. Which one of the following diagrams shows the cor-rect lines of force?

(A)

+

(B)

+

(C)

+

–

(D)

–

+

51. If the electric field is uniform, then the electric lines of forces are:

(A) Divergent (B) Convergent (C) Circular (D) Parallel

52. Electric lines of forces: (A) Exist everywhere (B) Are imaginary (C) Exist only in the immediate vicinity of electric

charges (D) None of these

53. Figure shows the electric lines of force emerging from a charged body. If the electric fields at A and B are EA and EB are respectively. If the distance between A and B is r, then:

BAr

(A) EA > EB (B) EA < EB

(C) EA = EB (D) EA = (EB)/r2

54. Two parallel plates of infinite dimensions are uni-formly charged. The surface charge density on one is σA will on the other is σB, field intensity at point C will be:

A

D

A

B

+

– – – – – –

+ + + + + + +

CB σ

σ


1.22 Chapter 1

(A) Proportional to (σA - σB)

(B) Proportional to (σA + σB)

(C) Zero

(D) 2σA

Section C: Electrostatic Potential Energy, Question Based on Energy Conservation and Angular Momentum Conservation, Potential Energy for a System of Charged Particles

55. A point positive charge of Q′ units is moved round another point positive charge of Q units in circular path. If the radius of the circle r is the work done on the charge Q′ in making one complete revolution is:

(A) Q

r4 0π∈ (B)

QQ

r

′∈4 0π

(C) ′∈Q

r4 0π (D) 0

56. Three charges are placed as shown in figure if the elec-tric potential energy of system is zero, then Q : q will be:

–q –qQ

rr

(A) Q

q=−2

1 (B) = =

Q

q

2

1

(C) Q

q=−1

2 (D)

Q

q=

1

4

57. State which one of the following is correct?

(A) Joule = Coulomb × Volt

(B) Joule = Coulomb/Volt

(C) Joule = Volt/Ampere

(D) Joule = Volt × Ampere

58. The KE in electron Volt gained by a α-particle when it moves from rest at point where its potential is 70 to a point where potential is 50 volts, is:

(A) 20 eV (B) 20 MeV

(C) 40 eV (D) 40 MeV

59. Point charge (q) moves form point (P) to point (S) along the path PQRS as shown in figure in a uniform electric field E. Pointing co-parallel to the positive direction of the X-axis. The co-ordinates of the points P, Q, R and S are (a, b, 0), (2a, 0, 0), (a, -b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression:

Q

E

PS

R

(A) q E a (B) -q E a

(C) q E a 2 (D) qE [( ) ]2a 2 2+ b

60. In a uniform electric field, the potential is 10 V at the origin of coordinates, and 8 V at each of the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). The potential at the point (1, 1, 1) will be:

(A) 0 (B) 4 V (C) 8 V (D) 10 V

61. When a negative charge is released and moves in elec-tric field, it moves toward a position of:

(A) Lower electric potential and lower potential energy

(B) Lower electric potential and higher potential energy

(C) Higher electric potential and lower potential energy

(D) Higher electric potential and higher potential energy

62. Two identical particles of mass m carry a charge Q each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v. The closed distance of approach is:

(A) 1

4 0

2

πεQ

mv (B)

1

4

4

0

2

2πεQ

mv

(C) 1

4

2

0

2

2πεQ

mv (D)

1

4

3

0

2

2πεQ

mv

63. Figure shows equi-potential surfaces for a two charges system. At which of the labeled points point will an electron have the highest potential energy?

(A) Point A (B) Point B (C) Point C (D) Point D

CA

D

B

–q +q

64. A proton is projected with velocity 7.45 × 105 m/s towards another proton which is at rest. The minimum approach is:


electrostatics 1.23

(A) 10-12 m (B) 10-14 m (C) 10-10m (D) 10-8 m

Section D: Electric Potential, Relation Between E and V., Potential Due to Point Charge, Rod, Ring, Disc, Question Based on Electric Field Intensity, and Electric Potential, Questions Based on Electric Field Lines

65. The dimensions of potential difference are: (A) [ML2T-2Q-1] (B) [MLT-2Q-1] (C) [MT-2Q-2] (D) [ML2 T-1 Q-1]

66. 1 esu of potential is equal to: (A) 1/300 volt (B) 8 ×1010 volt (C) 300 volt (D) 3 volt

67. The earth’s surface is considered to be at: (A) Zero potential (B) Negative potential (C) Infinite potential (D) Positive potential

68. ABC is equilateral triangle of side 1 m. Charges are placed at its corners as shown in Figure. O is the mid- point of side BC the potential at point (O) is:

OB C

A–6 cμ

–3 cμ–2 cμ

(A) 2.7 × 103 V (B) 1.52 × 105 V

(C) 1.3 × 103 V (D) - 1.52 × 105 V

69. In a region where E = 0, the potential (V) varies with distance r as:

(A) Vr

∝1

(B) V r∝

(C) Vr

∝12

(D) V = Constant, independent of (r)

70. Charges of +10

3

× 10-9 are placed at each of the

four corners of a square of side 8 cm. The potential at the intersection of the diagonals is:

(A) 150 2 volt (B) 1500 2 volt

(C) 900 2 volt (D) 900 volt

71. Three charges 2q, -q, -q are located at the vertices of an equilateral triangle. At the circum-center of the triangle:

(A) The field is zero but potential is not zero. (B) The field is non-zero but the potential is zero. (C) Both, field and potential are zero. (D) Both, field and potential are non-zero.

72. Three equal charges are placed at the three corners of an isosceles triangle as shown in the figure. The state-ment which is true for electric potential V and the field intensity E at the centre of the triangle is:

O

qq

q

(A) V = 0, E = 0 (B) V = 0, E ≠ 0

(C) V ≠ 0, E =0 (D) V ≠ 0, E ≠ 0

73. The electric potential V at any point (x, y, z) in space is given by V = 4x2 volt. The electric field E in V/m at the point (1, 0, 2) is:

(A) +8 in x direction (B) 8 in -x direction (C) 16 in + x direction (D) 16 in -x direction

74. The electron potential (V) as a function of distance (x) [in meters] is given by V = (5x2 + 10x -9) V.

The value of electric field at x =1 m would be: (A) 20 V/m (B) 6 V/m

(C) 11 V/m (D) -23 V/m

75. A uniform electric field having a magnitude E0 and direction along positive X-axis exists. If the electric potential (V) is zero at x = 0 then its value at x = +x will be:

(A) Vx = xE0 (B) Vx = -x · E0

(C) Vx = x2E0 (D) Vx = -x2E0

76. When charge of 3 coulomb is placed in a Uniform electric field, it experiences a force of 3000 newton, within this field, potential difference between two points separated by a distance of 1 cm is:

(A) 10 V (B) 90 V

(C) 1000 V (D) 3000 V

77. The potential difference between points A and B in the given uniform electric field is:


1.24 Chapter 1

E

a

A E

C B

b

(A) Ea (B) E a b( )2 2+

(C) Eb (D) (Eb/ 2 )

78. An equipotential surface is that surface: (A) On which each and every point has the same

potential (B) Which has negative potential (C) Which has positive potential (D) Which has zero potential

79. Some equipotential lines are as shown is figure E1, E2 and E3 are the electric fields at points 1, 2 and 3 then:

21

70 V60 V

50 V 40 V30 V20 V

3

(A) E1 = E2 = E3 (B) E1 > E2 > E3

(C) E1 > E2, E2< E3 (D) E1 < E2 < E3

Section E: Electric Field and Electric Potential Due to Electric Dipole, Dipole in Electric Field (Torque, Potential Energy), Angular Shm for Dipole, Force on Electric Dipole in Non-uniform Electric Field

80. If an electric dipole is kept in a uniform electric field, Then it will experience:

(A) A force (B) A couple and mover (C) A couple and rotates (D) A force and moves

81. The ratio of the electric field due to an electric dipole on its axis and on the perpendicular bisector of the dipole is:

(A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1

82. The region surrounding a stationary electric dipole has:

(A) Electric field only (B) Magnetic field only (C) Both electric and magnetic fields (D) Neither electric nor magnetic field

83. An electric dipole consists of two opposite charges each of magnitude 1 × 10-6 C separated by a distance 2 cm. The dipole is placed in an external field of 10 × 105 N/C. The maximum torque on the dipole is:

(A) 0.2 × 10-3 N-m (B) 1.0 × 10-3 N-m (C) 2 × 10-3 N-m (D) 4 × 10-3 N-m

84. The electric potential at a point due to an electric dipole will be:

(A) kp r

r

�� ⋅3

(B) kp r

r

�� ⋅2

(C) k p r

r

( )�� ×

(D) k p r

r

( )�� ×2

85. A uniformly charged rod with charge per unit length λ is bent in to the shape of a semicircle of radius R. The electric field at the centre is:

(A) 2k

R

λ (B)

k

R

λ2

(C) Zero (D) None of these

86. Figure shows the electric field lines around an electric dipole. Which of the arrows best represents the electric field at point P?

P

+– +

(A) (B)

(C) (D)

Section F: Questions Based on Electric Flux for Various Configuration, Question on Solid Angle Concept, Gauss Law Based Questions

87. The surface of a conductor: (A) Is a non-equipotential surface (B) Has all the points at the same potential (C) Has different points at different potential (D) Has at least two points at the same potential

88. When no charge is confined with in the Gauss’s surface, it implies that:


electrostatics 1.25

R

E

(A) πR2E (B) 2πR2E

(C) 4πR2E (D) (πR2E)/2

97. A cylinder of radius (R) and length (L) is placed in a uniform electrical field (E) parallel to the axis of the cyclinder. The total flux for the surface of the cylinder is given by:

(A) 2πR2E (B) πR2E

(C) π πR R

E

2 2+ (D) Zero

98. A surface enclosed an electric dipole, the flux through the surface is:

(A) Infinite (B) Positive (C) Negative (D) Zero

99. A square of side 20 cm is enclosed by a surface of sphere of 80 cm radius. Square and sphere have the same centre. Four charges +2 × 10-6 C, -5 × 10-6 C, -3 × 10-6 C, + 6 × 10-6 C are located at the four corners of a square, Then out going total flux from spherical surface in N-m2/C will be:

(A) Zero (B) (16π) × 10-6 (C) (8π) × 10-6 (D) (36 π) × 10-6

100. A charge (q) is located at the centre of a cube. The electric flux through any face of the cube is:

(A) q

∈0

(B) q

2 0∈

(C) q

4 0∈ (D)

q

6 0∈

101. The volume charge density as a function of distance X from one face inside a unit cube is varying as shown in the figure. Then the total flux (in SI units) through the cube if:

(ρ0 = 8.85 × 10-12 C/m3) is:

density

1 (in m)

ρ

1/4 3/4X

0

(A) 1/4 (B) 1/2 (C) 3/4 (D) 1

(A) E = 0

(B) E��

and ds��

are parallel

(C) E��

and ds��

are mutually perpendicular

(D) E��

and ds��

are inclined at some angle

89. If three electric di-poles are placed in some closed sur-face, then the electric flux emitting from the surface will be:

(A) Zero (B) Positive (C) Negative (D) None of these

90. If electric field flux coming out of a closed surface is zero, the electric field at the surface will be:

(A) Zero (B) Same at all places (C) Dependent upon the location of points (D) Infinites

91. The electric flux coming out of the equi-potential sur-face is:

(A) Perpendicular to the surface (B) Parallel to the surface (C) In all directions (D) Zero

92. A rectangular surface of 2 m width and 4 m length, is placed in an electric field of intensity 20 N/C, there is an angle of 60° between the perpendicular to surface and electrical field intensity. Then total flux emitted from the surface will be: (In Volt-metre)

(A) 80 (B) 40 (C) 20 (D) 160

93. A charge of Q coloumb is located at the centre of a cube. If the corner of the cube is taken as the origin, then the flux coming out from the faces of the cube in the direction of X-axis will be:

(A) 4πQ (B) Q/6∈0

(C) Q/3∈0 (D) Q/4∈0

94. A charge q is inside a closed surface and charge -q is outside. The out going electric flux is:

(A) -q/∈0 (B) Zero

(C) q/∈0 (D) 2q/∈0

95. For which of the following fields, Gauss’s law is valid: (A) Fields following square inverse law (B) Uniform field (C) All types of field (D) This law has no concern with the field

96. A hemisphere (radius R) is placed in electric field as shown in figure total outgoing flux is:


1.26 Chapter 1

Section G: Question Based on Equipotential Surface

102. An uncharged sphere of metal is placed in a uniform electric field produced by two large conducting par-allel plates having equal and opposite charges, then lines of force look like:

(A) +

– – – – – – –

+ + + + + + (B) +

– – – – –

+ + + +

(C) + + + + +

– – – – –

(D) + +

–

–

+++

–––

Section H: Properties of Conductors

103. A hollow sphere of charge does not produce an elec-tric field at any:

(A) Interior point (B) Outer point (C) Surface point (D) None of these

104. The earth had a net charge equivalent to 1 electron/m2 of surface area of radius 6.4 × 106 m. Its potential would be:

(A) +0.12 V (B) -0.12 V (C) +1.2 V (D) -1.2 V

105. The electric field inside a spherical shell of uniform surface charge density is:

(A) Zero (B) Constant, different from zero (C) Proportional to the distance from the centre (D) None of these

106. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at the centre of the sphere is:

(A) 0 V (B) 10 V (C) Same as at point 5 cm away from the surface (D) Same as at point 25 cm away from the surface

107. A large isolated metal sphere of radius (R) carries a fixed charge. A small charge is placed at a distance (r) from its surface experiences a force which is:

(A) Proportional to R (B) Independent of R and (C) Inversely proportional to (R + r)2

(D) Inversely proportional to r2

108. The electric field intensity at a point located at dis-tance r (r < R) from the center of a spherical conduc-tor (radius R) charged Q will be:

(A) kQR/r3 (B) kQr/R3

(C) kQ/r2 (D) Zero

109. The dependence of electric potential V on the distance r from the centre of a charged spherical shell is shown by:

(A) V

r

(B) V

r

(C) V

r

(D) V

r

110. Potential difference between centre and the surface of sphere of radius R and uniform volume charge density ρ within it will be:

(A) ρR2

06∈ (B)

ρR2

04∈

(C) 0 (D) ρR2

02∈

111. A solid sphere of radius R is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre?

(A) R (B) R/2 (C) R/3 (D) 2R

112. Two similar conducting spherical shells having charges 40 µC and -20 µC are some distance apart. Now they are touched and kept at same distance. The ratio of the initial to the final force between them is:

(A) 8 : 1 (B) 4 : 1 (C) 1 : 8 (D) 1 : 1

113. Three concentric conducting spherical shells carry charges as follows +4Q on the inner shell, -2Q on the middle shell and -5Q on the outer shell. The charge on the inner surface of the outer shell is:

(A) 0 (B) 4 Q (C) -Q (D) -2Q

114. A solid metallic sphere has a charge +3Q. Concentric with this sphere is a conducting spherical shell having charge -Q. The radius of the sphere is a and that of the spherical shell is b (>a). What is the electric field at a distance r (a < r < b) from the centre?


electrostatics 1.27

(A) 1

4 0πεQ

r (B)

1

4

3

0πεQ

r

(C) 1

4

3

02πεQ

r (D)

1

4 02πε

Q

r

115. Both question (a) and (b) refer to the system of charges as shown in the figure. A spherical shell with an inner radius a and an outer radius b is made of conducting material. A point charge +Q is placed at the centre of the spherical shell and a total charge -q is placed on the shell charge -q is distributed on the surfaces as:

Q a

b

–q

(A) -Q on the inner surface, -q on outer surface (B) -Q on the inner surface, -q +Q on the outer

surface

(C) +Q on the inner surface, -q - Q on the outer surface

(D) The charge -q is spread uniformly between the inner and outer surface

116. In the previous question assume that the electrostatic potential is zero at an infinite distance from the spheri-cal shell. The electrostatic potential at a distance R(a < R

< b) from the centre of the shell is: (where K =1

4 0πε)

(A) 0 (B) KQ

a

(C) KQ q

R

− (D) K

Q q

b

−

117. A positive charge q is placed in a spherical cavity made in a positively charged sphere. The centres of sphere and cavity are displaced by a small distance

�l.

Force on charge q is:

(A) In the direction parallel to vector �l

(B) In radial direction

(C) In a direction which depends on the magnitude of charge density in sphere

(D) Direction cannot be determined

LeveL ii

(A) 16.6 cm (B) 22.3 cm (C) 35.0 cm (D) 28.4 cm

4. Which of the following charge cannot present on oil drop in Millikan’s experiment?

(A) 4.0 × 10-19 C (B) 6.0 × 10-19 C (C) 10.0 × 10-19 C (D) All of these

5. If in Millikan’s oil drop experiment charges on drops are found to be 8 µC, 12 µC, 20 µC, then quanta of charge is:

(A) 8 µC (B) 4 µC (C) 20 µC (D) 12 µC

6. Two charges are placed as shown in figure Where should be a third charge be placed so that it remains in rest condition:

70 cm

9e 16e

(A) 30 cm from 9e (B) 40 cm from 16e

(C) 40 cm from 9e (D) (A) or (B)

Section A: Coloumb’s Law, Electrostatic Equillibrium + Properties of Charge

1. Determine dimensions of ε0 (permitivity of free space):

(A) [M-1L-3T4A2] (B) [M-1L-3T2A4]

(C) [ML3T-4A-2] (D) [M-1L-3A2T2]

2. Which of following result gives correct relation between charge induced (q) and inducing charge (q0)?

(A) q qr

0 11

= −∈

(B) q q

r

= −∈

0 1

1

(C) q qr

+ =∈01

(D) q q r= −∈0 1( )

3. Two point charge q1 and q2 are placed at a distance of 50 cm from each other in air, and interact with a certain force. Now the same charges are put in an oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now is:


1.28 Chapter 1

7. Which one of the following pattern of electric line of force cannot possible?

(A) (B)

(C) (D)

8. Two balls carrying charges +7 µC and -5 µC attract each other with a force F. If a charge -2 µC is added to both, the force between them will be:

(A) F (B) F

2

(C) 2F (D) Zero

9. The dielectric constant of a metal is: (A) ∞ (B) 0 (C) 1 (D) None of these

10. Two charges of equal magnitude q are placed at a dis-tance 2a. Another charge q of mass m, is placed mid-way between the two charges on X-axis. If this charges is displaced from equilibrium state to a distance x(x << a), then the particle:

(A) Will execute simple harmonic motion about equi-librium position

(B) Will be oscillating about equilibrium position but will not execute simple harmonic motion

(C) Will not return back to the equilibrium position (D) Will stop at equilibrium position

11. Two equal and like charges when placed 5 cm apart experience a repulsive force of 0.144 N. The magni-tude of the charge in micro-coulomb will be:

(A) 0.2 (B) 2 (C) 20 (D) 12

12. You have 1023 carbon atoms imagine that all the nuclei are put at the north pole of the earth and the electron at the south pole of the earth (radius = 6400 km.) Then force between the charges is: (Approximately)

(A) 5 × 105 N (B) 2 × 106 N (C) 13.6 × 103 N (D) 2 × 105 N

13. In 1 g of a solid, there are 5 × 1021 atoms. If one elec-tron is removed from everyone of 0.01% atoms of the solid, the charge gained by the solid is:

(Electronic charge is 1.6 × 10-19 C) (A) +0.08 C (B) +0.8 C (C) -0.08 C (D) -0.8 C

14. Two point charges of +2 µC and +6 µC repel each other with a force of 12 N. If each is given an addi-tional charge of -4 µC, then force will become:

(A) 4 N (attractive) (B) 60 N (attractive) (C) 4 N (repulsive) (D) 12 N (attractive)

15. What equal charges would to be placed on earth and moon to neutralize their gravitational attraction? (Use mass of earth = 1025 kg, mass of moon = 1023 kg)

(A) 8.6 × 1013 C (B) 6.8 × 1026 C

(C) 8.6 × 103 C (D) 9 × 106 C

16. Two identical small spheres carry charge of Q1 and Q2 with Q1>>Q2. The charges are d distance apart. The force they exert on one another is F1. The spheres are made to touch one another and then separated to dis-tance d apart. The force they exert on one another now is F2. Then F1/F2 is:

(A) 4 1

2

Q

Q (B)

Q

Q1

24

(C) 4 2

1

Q

Q (D)

Q

Q2

14

Section B: Electric Field, Questions on Constant Electric Field, Electric Field Due to Uniformly Charged Rod, Time Period of Shm + Electrostatics, Electric Field Due to Uniformly Charged Ring, Disc, Sheet

17. A sphere of radius R, is charged uniformly with total charge Q. Then correct statement for electric field is (r = distance from centre):

(A) KQr

R3, where r < R

(B) KQ

r2, where r ≥ R

(C) It is zero, at all points (D) (A) and (B) both

18. Choose the correct statements: (a) The tangent drawn at any point on the line of force

gives the direction of the force acting on a positive charge at that point.

(b) The normal drawn at any point on the line of force gives the direction of the force acting on a positive charge at that point.

(c) Electric lines of force start from a negative charge and end on a positive charge.

(d) Electric lines of force start from a positive charge and end on a negative charge.

(A) a, c (B) b, d (C) a, d (D) b, c


electrostatics 1.29

19. Choose the correct statements: (i) The density of electric lines of force at a point is

independent of the magnitude of electric intensity vector E at that point.

(ii) The density of electric lines of force at a point is proportional to the magnitude of electric intensity vector E at that point.

(iii) Actually, the electric field lines do not exist. This is just a graphical description of the electric field.

(iv) Actually, the electric field lines exist. (A) (ii), (iii) (B) (iii), (iv) (C) (ii), (iv) (D) (i), (iv)

20. A ring of radius R is charged uniformly with a charge +Q. The electric field at any point on its axis at a dis-tance r from the circumference of the ring will be:

(A) KQ

r (B)

KQ

r2

(C) KQ

rr R

32 2 1 2( ) /− (D)

KQr

R3

21. Two positive charges of 1 µC and 2 µC are placed 1 m apart. The value of electric field in N/C at the middle point of the line joining the charges will be:

(A) 10.8 × 104 (B) 3.6 × 104

(C) 1.8 × 104 (D) 5.4 × 104

22. The electric field in a certain region is given by

EK

xi

��=

3

ˆ. The dimensions of K are:

(A) [MLT-3A-1] (B) [ML-2T-3A-1]

(C) [ML4T-3A-1] (D) Dimensionless

23. Two infinite linear charges are placed parallel to each other at a distance 0.1 m from each other. If the linear charge density on each is 5 µC/m, then the force acting on a unit length of each linear charge will be:

(A) 2.5 N/m (B) 3.25 N/m (C) 4.5 N/m (D) 7.5 N/m

24. Figure shows field lines of an electric field, the line spacing parallel to the page is same everywhere. If the magnitude of the field at A is 40 N/C, then the magni-tude of the field at B is approximately:

B

A

yx

≈(y 2x)

(A) 40 N/C (B) 80 N/C (C) 20 N/C (D) Cannot be determined

25. Semicircular ring of radius 0.5 m. is uniformly charged with a total charge of 1.4 × 10-9 C. The electric field intensity at centre of this ring is:

(A) Zero (B) 320 V/m (C) 64 V/m (D) 32 V/m

26. A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of same mass is attached to the other end of the rod. The two particles carry charges +q and q respectively. This arrangement is held in a region of uniform electric field E such that the rod makes a small angle θ(< 50) with the field direction. The minimum time needed for the rod to become par-allel to the field after it is set free is: (Rod rotates about centre of mass)

(A) 22

πML

qE⋅ (B) π

ML

qE2⋅

(C) π2 2

ML

qE⋅ (D) 4

2π

ML

qE⋅

Section C: Electrostatic Potential Energy, Question Based on Energy Conservation and Angular Momentum Conservation, Potential Energy for a System of Charged Particles

27. A charge q = 10-6 C of mass 2 g (figure) is free to move then calculate its speed, when it is at a distance of b:

[Assume a = 1 m, b = 10 m, Q = 10-3 C]

fixed b

Q a q

(A) 90 m/s (B) 9 m/s (C) 900 m/s (D) None of these

28. Three charges q, 2q and 8q are to be placed on a 9 cm long straight line. Where the charges should be placed so that the potential energy of this system is minimum?

(A) q charge between 2q and 8q charges and 3 cm from charge 2q

(B) q charge between 2q and 8q charges and 5 cm from the charge 2q

(C) 2q charge between q and 8q charges and 7 cm from the charge q

(D) 2q charge between q and 8q charges and 9 cm from the charge q


1.30 Chapter 1

29. In the electric field of charge Q, another charge is car-ried from A to B, A to C, A to D and A to E, then work done will be:

Q

AB

⊕

C

DE

(A) Minimum along path AB (B) Minimum along path AD (C) Minimum along path AE (D) Zero along all the paths

30. Choose the incorrect statement: (A) The potential energy per unit positive charge in

an electric field at some point is called the electric potential.

(B) The work required to be done to move a point charge from one point to another in an electric field depends on the position of the points.

(C) The potential energy of the system will increase if a positive charge is moved against of Colombian force.

(D) The value of fundamental charge is not equivalent to the electronic charge.

31. A charge 10 esu is placed at a distance of 2 cm. from a charge 40 esu and 4 cm. from another charge -20 esu. The potential energy of the charge 10 esu is: (In ergs)

(A) 87.5 (B) 112.5 (C) 150 (D) Zero

32. A point charge q of mass m is located at the centre of a ring having radius R and charge Q. When it is displaced slightly, the point charge accelerates along the X-axis to infinity, the ultimate speed of the point charge:

(A) 2kQq

mR (B)

kQq

mR

(C) kQq

mR2 (D) Zero

Section D : Electric Potential, Relation between E and V., Potential Due to Point Charge, Rod, Ring, Disc, Question Based on Electric Field Intensity, and Electric Potential, Questions Based on Electric Field Lines

33. Four charges 2 C, -3 C, -4 C and 5 C respectively are placed at all the corners of a square. Which of the

following statements is true for the point of intersec-tion of the diagonals:

(A) E = 0, V = 0 (B) E ≠ 0, V = 0 (C) E = 0, V ≠ 0 (D) E ≠ 0, V ≠ 0

34. Which statement is true? (i) A ring of radius R carries a uniformly distributed

charge +Q. A point charge -q is placed on the axis of the ring at a distance 2R from the centre of the ring and released from rest. The particle executes a simple harmonic motion along the axis of the ring.

(ii) Electrons move from a region of higher potential to that of lower potential.

(A) Only (i) (B) Only (ii) (C) (i), (ii) (D) None of these

35. A non-conducting ring of radius 0.5 m, 1.11 × 10-10 coulomb charge is non-uniformly distributed over the circumference of ring, produces electric field E around itself. If l = 0 is the centre of ring, then the value of

− ⋅=−∞∫ E d l

l

0

is:

(A) 2 V (B) - 2 V (C) - 1 V (D) Zero

36. A spherical droplet having a potential of 2.5 volt is obtained as a result of merging of 125 identical droplets. Find the potential of the constituent droplet:

(A) 0.4 V (B) 0.5 V (C) 62.5 V (D) 0.1 V

37. 15 joule of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. Then the potential difference (VB - VA) is:

(A) 1500 volt (B) -1500 volt (C) 0.15 volt (D) None of these

Section E: Electric Field and Electric Potential Due to Electric Dipole, Dipole in Electric Field (Torque, Potential Energy), Angular Shm for Dipole, Force on Electric Dipole in Non-uniform Electric Field

38. At any point on the right bisector of line joining two equal and opposite charges:

(A) The electric field is zero (B) The electric potential is zero (C) The electric potential decreases with increasing

distance from centre (D) The electric field is perpendicular to the line join-

ing the charges


electrostatics 1.31

39. For a dipole, the value of each charge is 10-10 state coulomb and separation is 1 Å, then its dipole moment is:

(A) One debye (B) 2 debye (C) 10-3 debye (D) 3 × 10-20 debye

Section F: Questions Based on Electric Flux for Various Configuration, Question on Solid Angle Concept, Gauss Law Based Questions

40. A sphere of radius R and charge Q is placed inside an imaginary sphere of radius 2R whose centre coincides with the given sphere. The flux related to imaginary sphere is:

(A) Q

∈0

(B) Q

2 0∈

(C) 4

0

Q

∈ (D)

2

0

Q

∈

41. 20 µC charge is placed inside a closed surface then flux related to surface is f. If 80 µC charge is added inside the surface then change in flux is:

(A) 4f (B) 5f (C) f (D) 8f

42. In a region of space the electric field is given by�E i j k= + +8 4 6ˆ ˆ ˆ. The electric flux through a surface of area of 100 units in XY-plane is:

(A) 800 units (B) 300 units (C) 400 units (D) 1500 units

43. The electric field in a region of space is given by E =

(5i + 2 j ) N/C. The electric flux due to this field through an area 2m2 lying in the YZ plane, in SI units is:

(A) 10 (B) 20

(C) 10 2 (D) 2 29

44. The total flux associated with given cube will be where a is side of cube.

1

4 9 100

9

∈= × ×

π

a

μ2 C

μ4 C

μ6 C

μ3 C

μ5 C

μ7 Cμ8 C

μ1 Ca

a

(A) 162π × 10-3 Nm2/C (B) 162π × 103 Nm2/C (C) 162π × 10-6 Nm2/C (D) 162π × 106 Nm2/C

Section H: Properties of Conductors

45. Charge Q distributed on two concentric metallic shells of radii r and R in such a way that their surface charge densities are the same. Electric potential at their com-mon centre would be:

(A) Q r R

R

( )+

∈4 02π

(B) Q r R

r R

( )

( )

+

+ ∈4 2 20π

(C) Q

r R4 2 2( )+ (D)

Q r R( )+∈4 0π

46. Force between two identical spheres charged with same charge is F. If 50% charge of one sphere is trans-ferred to second sphere then new force will be:

(A) 3

4F (B)

3

8F

(C) 3

2F (D) None of these

47. A solid conducting sphere having a charge Q is sur-rounded by an uncharged concentric conducting spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the shell be V. If the shell is now given a charge of -3Q the new potential difference between the same two surfaces is:

(A) V (B) 2V (C) 4V (D) -2V

48. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre.

(i) Increases as r increases, for r < R (ii) Decreases as r increases, for 0 < r < ∞ (iii) Decreases as r increases, for R < r < ∞ (iv) Is discontinuous at r = R (A) (i), (iii) (B) (iii), (iv) (C) (i), (ii) (D) (ii), (iv)

49. Two conductors of the same shape and size. One of copper and the other of aluminium (less conducting) are placed in an uniform electric field. The charge induced in aluminium:

(A) Will be less than in copper (B) Will be more than in copper (C) Will be equal to that in copper (D) Will not be connected with copper


1.32 Chapter 1

50. A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to a potential 3 esu. When the outer sphere is earthed, the charge on the inner sphere is:

(A) 54 esu (B) 1

4esu

(C) 30 esu (D) 36 esu

Direction for Questions 51 to 64: (A) If both Statement I and Statement II are true,

and Statement II is the correct explanation of Statement I.

(B) If both Statement I and Statement II are true but Statement II is not the correct explanation of Statement I.

(C) If Statement I is true but Statement II is false. (D) If Statement I is false but Statement II is true.

51. Statement I: Charge is invariant. Statement II: Charge does not depends on speed or

frame of reference.

52. Statement I: Mass of ion is slightly differed from its element.

Statement II: Ion is formed, when some electrons are removed or added so mass changes.

53. Statement I: Total number of positive ions in nature is constant.

Statement II: Charge is conserved.

54. Statement I: If a positive is charge free to move then it moves from higher potential to lower potential.

Statement II: Force on positive charge is along �E,

which is HP to LP.

55. Statement I: A point charge Q is rotated in a circle of radius r around a charge q. The work done will be zero.

Statement II: For this motion the force is along the radius and direction of motion is perpendicular.

56. Statement I: Charge is quantized. Statement II: Charge, which is less than 1 C is not possible

57. Statement I: Electric lines are always straight and continuous.

Statement II:Electric lines represents velocity field.

58. Statement I: Electric potential energy of any positive charge is always positive.

Statement II: Potential energy is a vector quantity.

59. Statement I: Quantization of charge is invalid after the presence of quark particles.

Statement II: Quark particle has charge greater than electron.

60. Statement I: In electrostatic electric lines of force can never be closed loops, as a line can never start and end on the same charge.

Statement II: The number of electric lines of force originating or terminating on a charge is proportional to the magnitude of charge.

61. Statement I: If a point charge q is placed in front of an infinite grounded conducting plane surface, the point charge will experience a force.

Statement II: This force is due to the induced charge on the conducting surface which is at zero potential.

62. Statement I: When two charged spheres are touched, then total charge is always devides equally.

Statement II: Induced charge always equal to induc-ing charge.

63. Statement I: If a charge enters in electric field then it moves in the direction of

�E.

Statement II: Force on charge is in the direction of field.

64. Statement I: The coulomb force is the strongest force in the universe.

Statement II: The coulomb force is weaker than the gravitational force.

LeveL iii A third charge q3 is moved along the arc of a circle of

radius 40 cm from C to D. The change in the potential

energy of the system is q3

04π ∈k, where k is:

[AIPMT, 2005]

(A) 8q2 (B) 6q2 (C) 8q1 (D) 6q1

2. As per this diagram a point charge +q is placed at the origin O. Work done in taking another point charge -Q

previous Years’ Questions—neet

1. Two charges q1 and q2 are placed 30 cm apart, as shown in the figure.

40cm

Cq3

q2q1

A B30cm


electrostatics 1.33

from the point A [coordinates (0, a)] to another point B [coordinates(a, 0)] along the straight path AB is: [AIPMT, 2005]

y

O B

A

X

(A) −∈

qQ

a4

12

02π

a

(B) Zero

(C) qQ

a4

1 1

202π ∈

(D) qQ

a4

12

02π ∈

a

3. Two infinitely long parallel conducting plates having surface charge densities +σ and -σ respectively are separated by a small distance. The medium between the plates is vacuum. If ∈0 is the dielectric permittivity of vacuum, then the electric field in the region between the plates is: [AIIMS, 2005]

(A) 0 volt/m (B) σ

2 0∈volt/m

(C) σ∈0

volt/m (D) 2

0

σ∈

volt/m

4. Two concentric conducting thin spherical shells A, and B having radii rA and rB(rB > rA) are charged to QA

and -QB (|QB| > |QA|). The electrical field along a line, (passing through the centre) is: [AIIMS, 2005]

(A) E

XrArB

0

(B) E

XrA0 rB

(C) E

XrA rB0

(D) E

XrA0

rB

5. Two infinitely long parallel conducting plates having surface charge densities +σ and -σ respectively, are separated by a small distance. The medium between the plates is vacuum. If ε0 is the dielectric permittivity of vaccum, then the electric field in the region between the plates is: [AIIMS, 2005]

(A) 0 (B) σ/2ε0 V/m (C) σ/2ε0 V/m (D) zσ/ε0 V/m

6. A square surface of side L metre is in the plane of the paper. A uniform electric field

�E (volt/m), also in the

plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is: [AIPMT, 2006]

E

(A) Zero (B) EL2

(C) EL2

02∈ (D)

EL2

2

7. An electric dipole of dipole moment �p is lying along

a uniform electric field �E. The work done in rotating

the dipole by 90° is: [AIPMT, 2006] (A) 2pE (B) pE

(C) 2 pE (D) pE

2

8. Two parallel large thin metal sheets have equal surface charge densitiesn (σ = 26.4 × 10-12 C/m2) of oppo-site signs. The electric field between these sheets is: [AIIMS, 2006]

(A) 1.5 N/C (B) 1.5 × 10-10 N/C (C) 3 N/C (D) 3 × 10-10 N/C

9. The spatial distribution of the electric field due to charges (AB) is shown in figure. Which one of the fol-lowing statements is correct? [AIIMS, 2006]

A B

(A) A is positive and B negative and |A|>|B| (B) A is negative and B positive |A| = |B| (C) Both are positive but A > B (D) Both are negative but A > B

10. The voltage of clouds is 4 × 106 V with respect to ground. In a lightning strike lasting 100 ms, a charge of 4 C is delivered to the ground. The power of light-ning strike is: [AIIMS, 2006]

(A) 160 MW (B) 80 MW (C) 20 MW (D) 500 kW


1.34 Chapter 1

11. Charges +q and -q are placed at points A and B respec-tively which are a distance 2L apart, C is the mid-point between A and B. The work done in moving a charge +Q along the semicircle CRD is: [AIPMT, 2007]

A C B D

(A) −∈

qQ

L6 0π (B)

qQ

L4 0π ∈

(C) qQ

L2 0π ∈ (D)

qQ

L6 0π ∈

12. A hollow cylinder has a charge q coulomb within it. If f is the electric flux in units of voltmeter associ-ated with the curved surface B, the flux linked with the plane surface A in units of voltmeter will be: [AIPMT, 2007]

C A

B

(A) q

∈−

0

φ (B) 1

2 0

q

∈−

φ

(C) q

2 0∈ (D)

φ3

13. Three point charges +q, -2q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0) respectively. The magnitude and direc-tion of the electric dipole moment vector of this charge assembly are: [AIPMT, 2007]

(A) 2 qa along +x direction

(B) 2 qa along +y direction

(C) 2 qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)

(D) qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)

14. Assertion (A): The lightning conductor at the top of high building has sharp pointed ends. [AIIMS, 2007]

Reason (R): The surface density of charge at sharp points is very high resulting in setting up of electric wind.

(A) Both A and R are correct and R is the correct explanation of A.

(B) Both A and R are correct but R is not the correct explanation of A.

(C) A is correct but R is incorrect. (D) Both A and R are incorrect.

15. The electric potential at a point in free space due to a charge Q coulomb is Q × 1011 V. The electric field at that point is: [CBSE AIPMT, 2008]

(A) 4πε0Q × 1022 V/m

(B) 12πε0Q × 1020 V/m

(C) 4πε0Q × 1020 V/m

(D) 12πε0Q × 1022 V/m

16. A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is: [CBSE AIPMT, 2008]

OC

A

K

B

D

(A) 3E along KO (B) E along OK (C) E along KO (D) 3E along OK

17. Charge q is uniformly distributed over a thin half ring of radius R. The electric field at the centre of the ring is: [AIIMS, 2008]

(A) q

R2 20

2π ε (B)

q

R4 20

2π ε

(C) q

R4 02πε

(D) q

R2 02πε

18. The electric potential at a point (x, y, z) is given by: [AIPMT, 2009]

V = -x2y - xz3 + 4 The electric field at that point is:

(A) �E = i (2xy - z3) + j xy2 + k 3z2x

(B) �E = i (2xy + z3) + j x2 + k 3xz2

(C) �E = i 2xy + j (x2 + y2) + k (3xz - y2)

(D) �E = i z3 + j xyz + k z2

19. Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, -σ and σ respectively. If VA, VB and VC denote the potentials of the three shells, then, for c = a + b, we have: [AIPMT, 2009]


electrostatics 1.35

(A) VC = VB = VA (B) VC = VA ≠ VB

(C) VC = VB ≠ VA (D) VC ≠ VB ≠ VA

20. Two positive ions, each carrying a charge q are sep-arated by a distance d. If F is the force of repulsion between the ions, then the number of electrons miss-ing from each ion will be: (e being the charge on an electron) [CBSE AIPMT, 2010]

(A) 4πε0

2Fd

e (B)

4πε02Fe

e

(C) 4πε0

2

2

Fd

e (D)

4 0

2

2

πε Fd

e

21. Four electric charges +q, +q, -q and -q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, mid-way between the two charges +q and +q is: [CBSE AIPMT, 2011]

–q

–q2L +q

2L+q

A

L

L

(A) 1

4

21

1

50πεq

L+

(B) 1

4

21

1

50πεq

L−

(C) Zero

(D) 1

4

21 5

0πεq

L( )+

22. An electric dipole of moment p is placed in an elec-tric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ = 90°, the torque and the potential energy of the dipole will respectively be: [CBSE AIPMT, 2012]

(A) ρ θ ρ θE Esin , cos2

(B) ρ θ ρ θE Ecos , sin−

(C) ρ θ ρ θE Esin , cos−

(D) ρ θ ρ θE Esin , cos−2

23. Assertion (A): If the bob of a simple pendulum kept in a horizontal electric field, its period of oscillation will remain same.

Reason (R): If bob is charged and kept in horizontal electric field, then the time period will be decreased. [AIIMS, 2012]

(A) Both A and R are correct and R is the correct explanation of A.

(B) Both A and R correct but R is not the correct explanation of A.

(C) A is correct but R is incorrect. (D) A is incorrect but R is correct.

24. Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10-2 C and 5 × 10-2 C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is: [CBSE AIPMT, 2012]

(A) 2 × 10-2 C (B) 3 × 10-2 C (C) 4 × 10-2 C (D) 1 × 10-2 C

25. Four point charges -Q, -q, 2q and 2Q are placed. one at each corner of the square. The relation between Q and q for which the potential at the centre of the square. [CBSE AIPMT, 2012]

(A) Q = -q (B) Q = -1

q

(C) Q = q (D) Q =1

q

26. Identify the incorrect statements for electric charge q: [AIIMS, 2012]

(A) Quantized (B) Conserved (C) Additive (D) Non-transferable

27. An electron is projected with velocity �v v x= 0 ˆ in an

electric field �E E y= 0 ˆ. Trace the path followed by the

electron: [AIIMS, 2012] (A) Parabola (B) Circle (C) Straight line in +y direction (D) Straight line in -y direction

28. A dipole of dipole moment p is placed in a non-uniform electric field along X-axis. Electric field is increas-ing at the rate of 1 V/m2 then the force on dipole is: [AIIMS, 2013]

(A) 0 (B) 2p (C) p/2 (D) p

29. Electric field at a distance r from an infinitely large conducting sheet is proportional to: [AIIMS, 2013]


1.36 Chapter 1

(A) r-1 (B) r-2

(C) r3/2 (D) r°

30. A conducting sphere of radius R is given a charge Q. The electric potential and electric fi eld at the centre of the sphere respectively are: [AIPMT, 2014]

(A) Zero and Q

R4 02πε

(B) Q

R4 0πεand zero

(C) Q

R

Q

R4 40 02πε πε

and

(D) Both are zero

31. In a region, the potential is represented by V(x, y, z) = 6x -8xy - 8y + 6yz, where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulombs situated at the point (1, 1, 1) is: [AIPMT, 2014]

(A) 6 5 N (B) 30 N

(C) 24 N (D) 4 35 N

32. Electric potential (f) of a quadrupole varies with dis-tance r on its axis as: [AIIMS, 2014]

(A) f ∝ r-1

(B) f ∝ r-2

(C) f ∝ r-3

(D) f ∝ r-3/2

33. The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius a centred at the origin of the fi eld, will be given by: [AIPMT, 2015]

(A) 4π∈0 Aa3 (B) ∈0Aa3

(C) 4π∈0 Aa3 (D) A∈0 a2

34. If potential (in volts) in a region is expressed as V(x, y, z) = 6xy - y + 2yz, the electric fi eld (in N/C) at point(1, 1, 0) is: [AIPMT, 2015]

(A) − + +( )6 9ˆ ˆ î j k (B) − + +( )3 5 3ˆ ˆ î j k

(C) − + +( )6 5 2ˆ ˆ î j k (D) − + +( )2 3ˆ ˆ î j k

35. A non-conducting spherical shell of diameter 10 cm has a charge of 1.6 × 10-4 C. A charge of 20 C is placed at a distance of 10 cm from its centre, then force between them will be: [AIIMS, 2015]

(A) 4 × 109 N (B) 16 × 1013 N (C) 3 × 109 N (D) 6 × 10-9 N

36. A spherical conducting shell of radius r0 carry a charge q0. The value of electric fi eld inside it is: [AIIMS, 2015]

(A) kq0/r2

(B) kq0/r3

(C) Zero

(D) Uniform but non zero

37. Charge of 5 µC each are placed at the corners of an equilateral triangle of side 10 cm. Then the force on each charge is: [AIIMS, 2015]

(A) 78 N (B) 39 N (C) 29 N (D) 22.5 N

38. Two identical charged spheres suspended from a com-mon point by two massless strings of lengths l, are ini-tially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity υ. Then υ varies as a function of the distance x between the spheres, as: [AIPMT, 2016]

(A) υ ∝ x1

2 (B) υ ∝ x

(C) υ ∝−

x1

2 (D) υ ∝ −x 1

39. The fi gures below show regions of equipotentials:

20 V 20 V20 V

AAAA BBBB

40 V 40 V 10 V 30 V

10 V(a) (b) (c) (d)

10 V30 V 30 V 20 V 40 V

40 V

30 V10 V

A positive charge is moved from A to B in each diagram, then: [NEET, 2017]

(A) In all the four cases the work done is the same (B) Minimum work is required to move q in fi gure (a) (C) Maximum work is required to move q in fi gure (b) (D) Maximum work is required to move q in fi gure (c)


electrostatics 1.37

previous Years’ Questions—Jee main

Electrostatics i (A) 2.65 × 106 m/s (B) 7.02 × 1012 m/s

(C) 1.87 × 106 m/s (D) 32 × 10-19 m/s

5. An electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience: [AIEEE, 2006]

(A) A translational force only in the direction of the field.

(B) A translational force only in a direction normal to the direction of the field.

(C) A torque as well as a translational force. (D) A torque only

6. The potential at a point x (measured in µm) due to some charges situated on the X-axis is given by

V x x( ) ( )= −20 42/ volt. [AIEEE, 2007]

The electric field E at x = 4 µm is given by:

(A) 5

3V m/µ ⋅ and in the -ve x-direction

(B) 5

3V m/µ ⋅ and in the +ve x-direction

(C) 10

9V m/µ ⋅ and in the -ve x-direction

(D) 10

9V m/µ ⋅ and in the +ve x-direction

7. Charges are placed on the vertices of a square as shown.

Let E��

be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then: [AIEEE, 2007]

A B

CD

q

–q –q

q

(A) E��

remains unchanged, V changes

(B) Both E��

and V change

(C) E��

and V remain unchanged

(D) E��

change, V remains unchanged

1. Two point charges +8q and -2q are located at x = 0 and x = L respectively. The location of a point on the X-axis at which the net electric field due to these two point charges is zero, is: [AIEEE, 2005]

(A) 2L (B) L/4 (C) 8L (D) 4L

2. Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and -q. The poten-tial difference between the centres of the two rings is: [AIEEE, 2005]

(A) qR

d4 02πε

(B) q

R R d2

1 1

02 2πε

−+

(C) Zero (D) q

R R d4

1 1

02 2πε

−+

3. A charged ball B hangs from a silk thread S. Which makes an angle θ with a large charged conducting sheet P, as shown in the figure. The surface charge density σ of the sheet is proportional to: [AIEEE, 2005]

PS

B

+

+

+

+

+

+

+

θ

(A) cosθ (B) cotθ (C) sinθ (D) tanθ

4. Two insulating plates are both uniformly charged in such a way that the potential difference between them is V V2 1 20− = V. (i.e., plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? [AIEEE, 2006]

(e = 1.6 × 10-19 C, m0 = 9.11 × 10-31 kg)

1 2

X

Y

0.1 m


1.38 Chapter 1

8. An electric charge 10 3− µC is placed at the origin (0, 0) of X-Y coordinate system. Two points A and B are

situated at ( , )2 2 and (2, 0) the respectively. The potential difference between the points A and B will be: [AIEEE, 2007]

(A) 9 V (B) Zero (C) 2 V (D) 4.5 V

9. ∆V measured between B and C is: [AIEEE, 2008]

(A) ρπα

ρπ α

I I

b−

+( )

(B) ρα

ρα

I I

b−

+( )

(C) ρπα

ρπ α

I I

b2 2−

+( )

(D) ρ

π αI

b2 ( )−

10. The questions contains Statements I and Statement II of the four choice given after the statements, choose the one that best describes the two statements.

Statements I: For a charged particle moving from point P to point Q, the net work done by an electro-static field on the particle is independent of the path connecting point P to point Q.

Statements II: The net work done by a conservative force on an object moving along a closed loop is zero. [AIEEE, 2009]

(A) Statement I is true, Statement II is true. (B) Statement I is true, Statement II is true; Statement

II is the correct explanation of Statement I. (C) Statement I is true, Statement II is true; Statement

II is not the correct explanation of Statement I. (D) Statement I is false, Statement II is true.

11. Two points P and Q are maintained at the potentials of 10 V and -4 V respectively. The work done in moving 100 electrons from P to Q is: [AIEEE, 2009]

(A) − × −19 10 17 J

(B) 9 60 10 17. × − J

(C) − × −2 24 10 16. J

(D) 2 24 10 16. × − J

12. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then the Q

q equals: [AIEEE, 2009]

(A) −2 2 (B) -1

(C) 1 (D) −1

2

13. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field E

��

at the centre O is: [AIEEE, 2010]

O

j

i

(A) q

rj

4 20

2π εˆ (B) −

q

rj

4 20

2π εˆ

(C) −q

rj

2 20

2π εˆ (D)

q

rj

2 20

2π εˆ

14. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 g/cm, the angle remains the same. If density of the material of the sphere is 16 g/cm3, the dielectric con-stant of the liquid is: [AIEEE, 2010]

(A) 4 (B) 3

(C) 2 (D) 1

15. Two positive charges of magnitude q are placed at the ends of a side 1 of a square of side 2a. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is: [AIEEE, 2011]

(A) 1

4

21

1

50πεqQ

a−

(B) Zero

(C) 1

4

21

1

50πεqQ

a+

(D) 1

4

21

2

50πεqQ

a−

16. Two identical charged spheres suspended from a com-mon point by two massless strings of length l are ini-tially a distance d(d << l) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity v. Then as functions of dis-tance x between them: [AIEEE, 2011]


electrostatics 1.39

(A) v x∝ −1 (B) v x∝ 1 2/

(C) v x∝ (D) v x∝ −1 2/

17. Two charges, each equal to q, are kept at x = -a and x = a on the X-axis. A particle of mass m and charge q0 = q

2 is placed at the origin. If charges q0 is given a small

displacement y(y << a) along the Y-axis, the net force acting on the particle is proportional to: [JEE Main, 2013]

(A) y (B) -y

(C) 1

y (D) −

1

y

18. Assume that an electric field �E x i= 30 2ˆ exists in

space. Then the potential difference VA - VO, where VO is the potential at the origin and VA the potential at x = 2 m is: [JEE Main, 2014]

(A) -80 J (B) 80 J

(C) 120 J (D) -120 J

19. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞) on its surface. For this sphere the equipotential surfaces with

potentials 3

2

5

4

3

4 40 0 0 0V V V V

, , and have radius R1, R2,

R3 and R4 respectively. Then: [JEE Main, 2015] (A) R1 = 0 and R2 < (R4 - R3) (B) 2R < R4 (C) R1 = 0 and R2 > (R4 - R3) (D) R1 ≠ 0 and (R2 - R1) > (R4 - R3)

20. A long cylindrical shell carries positive surface charge σ in the upper half and negative surface charge -σ in the lower half. The electric field lines around the cylin-der will look like figure given in:

(Figures are schematic and not drawn to scale) [JEE Main, 2015]

(A)

–––––––––

+++++++++

(B)

–––––––––

+++++++++

(C)

–––––––––

+++++++++

(D)

–––––––––

+++++++++

Electrostatics II

21. Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uni-formly charged. If the spheres are connected by a con-ducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is: [AIEEE, 2006]

(A) 4 : 1 (B) 1 : 2

(C) 2 : 1 (D) 1 : 4

22. A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r) produced by the shell in the range 0 ≤ r < x, where r is the distance from the centre of the shell? [AIEEE, 2008]

(A)

O Rr

E(r) (B)

O Rr

E(r)

(C)

O Rr

E(r) (D)

O R r

E(r)

23. Let Q

Rr

π 4 ⋅ be the charge density distribution for a

solid sphere of radius R and total charge Q. For a point P inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is: [AIEEE, 2009]


1.40 Chapter 1

(A) Zero (B) Q

r4 0 12πε

(C) Qr

R12

044πε

(D) Qr

R12

043πε

24. Let there be a spherically symmetric charge distribu-

tion with charge density varying as ρ ρ( )rr

R= −

0

5

4

upto r = R, and ρ( )r = 0 for r > R, where r is the dis-tance from the origin. The electric field at a distance r (r < R) from the origin is given by: [AIEEE, 2010]

(A) 4

3

5

30

0

πρε

r r

R−

(B)

ρε0

04

5

3

r r

R−

(C) 4

3

5

40

0

ρε

r r

R−

(D)

ρε0

03

5

4

r r

R−

25. The electrostatic potential inside a charged spherical ball is given by φ = +ar b2 where r is the distance from the centre a, b are constants. Then the charge density inside the ball is: [AIEEE, 2011]

(A) −6 0a rε (B) −24 0π εa

(C) −6 0aε (D) −24 0π εa r

26. This question has statement 1 and statement 2 of the four choices given after the statements, choose the one that best describes the two statements. An insulating solid sphere of radius R has a uniform positive charge density ρ. As a result of this uniform charge distribu-tion, there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point outside the sphere. The electric potential at infinite is zero. [AIEEE, 2012]

Statement I: When a charge q is taken from the cen-tre of the surface of the sphere its potential energy

changes by qρε3 0

.

Statement II: The electric field at a distance r (r < R)

from the centre of the sphere is ρεr

3 0

.

(A) Statement I is false, Statement II is true. (B) Statement I is true, Statement II is false. (C) Statement I is true, Statement II is true; Statement

II is the correct explanation for Statement I. (D) Statement I is true, Statement II is true; Statement

II is not the correct explanation of Statement I.

27. In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as function of distance from the centre. The graph which would cor-respond to the above will be: [AIEEE, 2012]

(A) E

R r

(B) E

R r

(C) E

R r

(D) E

R r

28. A charge Q is uniformly distributed over a long rod AB of length l as shown in the figure. The electric potential at the point O lying at a distance L from the end A is: [JEE Main, 2013]

LAO B

L

(A) Q

L4 20πε ln (B) Q

L

ln 2

4 0πε

(C) Q

L8 0πε (D)

3

4 0

Q

Lπε

29. The region between two concentric spheres of radii a and b, respectively (see figure), has volume charge

density ρ =A

r, where A is a constant and r is the dis-

tance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be con-stant, is: [JEE Main, 2016]

Q

a

b

(A) Q

b a2 2 2π ( )− (B)

22 2

Q

a bπ ( )−

(C) 2

2

Q

aπ⋅ (D)

Q

a2 2π⋅


electrostatics 1.41

eXerCises

Level i

1. A 2. B 3. D 4. A 5. A 6. D 7. D 8. D 9. B 10. D 11. C 12. A 13. B 14. C 15. A 16. B 17. A 18. A 19. A 20. D 21. A 22. C 23. B 24. B 25. A 26. C 27. C 28. D 29. D 30. A 31. D 32. B 33. C 34. A 35. B 36. D 37. C 38. A 39. C 40. A 41. B 42. B 43. B 44. C 45. C 46. C 47. A 48. D 49. D 50. B 51. D 52. B 53. A 54. B 55. D 56. D 57. A 58. C 79. B 60. B 61. C 62. B 63. B 64. A 65. A 66. C 67. A 68. D 69. D 70. B 71. B 72. C 73. B 74. A 75. B 76. A 77. C 78. A 79. C 80. C 81. B 82. A 83. C 84. A 85. A 86. B 87. B 88. C 89. A 90. C 91. A 92. A 93. C 94. C 95. A 96. A 97. D 98. D 99. A 100. D 101. C 102. C 103. A 104. B 105. A 106. B 107. C 108. D 109. B 110. A 111. C 112. A 113. D 114. C 115. B 116. D 117. A

Level ii

1. A 2. B 3. B 4. D 5. B 6. D 7. C 8. A 9. A 10. A 11. A 12. A 13. A 14. A 15. A 16. C 17. D 18. C 19. A 20. C 21. B 22. C 23. C 24. C 25. D 26. C 27. A 28. A 29. D 30. D 31. C 32. A 33. B 34. D 35. A 36. D 37. A 38. B 39. A 40. A 41. A 42. B 43. A 44. B 45. B 46. A 47. A 48. A 49. C 50. D 51. A 52. A 53. D 54. A 55. A 56. C 57. D 58. D 59. D 60. B 61. A 62. D 63. D 64. D

Level iii

Previous Years’ Questions—NEET

1. A 2. B 3. C 4. A 5. C 6. A 7. B 8. C 9. A 10. A 11. A 12. B 13. C 14. A 15. A 16. B 17. A 18. B 19. B 20. C 21. B 22. C 23. A 24. B 25. A 26. D 27. A 28. D 29. D 30. B 31. D 32. B 33. A 34. C 35. C 36. C 37. B 38. C 39. A

answer keYs

30. An electric dipole has a fixed dipole moment �P,

which makes an angle θ with respect to X-axis. When

subjected to an electric field �E Ei1 = ˆ, it experiences

torque �T k1 =τ ˆ. When subjected to another electric

field �E E j2 13= ˆ, it experiences a torque

� �T T2 1= − .

The angle θ is: [JEE Main, 2017] (A) 90° (B) 30° (C) 45° (D) 60°


1.42 Chapter 1

electrostatic 1

Previous Years’ Questions—JEE Main

1. A 2. B 3. D 4. A 5. C 6. D 7. D 8. B 9. C 10. B 11. D 12. A 13. C 14. C 15. A 16. D 17. A 18. A 19. A, B 20. C 21. C 22. A 23. C 24. B 25. C 26. A 27. C 28. B 29. D 30. D


NEET PHY CH1 Module 3 - Baluni Classes

Documents

Transcript of NEET PHY CH1 Module 3 - Baluni Classes