Post on 07-Apr-2018
8/6/2019 Physics Impt Point
1/34
Mechanics
8/6/2019 Physics Impt Point
2/34
8/6/2019 Physics Impt Point
3/34
Graphical Representaion: Day 2
Constant Velocity. No change in velocity (either in magnitude or in direction). Thenvelocity = displacement / time
See the V-t and S-t graph
Slope of s-t graph gives velocity Area under V-t graph gives displacement
Constant Acceleration: Three equation of motion. See the S-t, V-t and V-s graph for various cases:
When initial velocity is zero
When initial velocity is not zero
Slope of S-t, V-t graph
Area under A-t and V-t graph
Solve cases where object undergoes motion in different segment with eachsegment having a different constant acceleration: We have to first identify suchsegments then calculate the individual segment separately.
Conversion of V-t graph into S-t graph and V-s graph
Calculation of distance and Displacement from area under V-t graph: For displacement: We take magnitude and sign
For distance: We take only magnitude
Analysis of impossible graphs: Time cannot be negative
Time cannot be decreasing
Distance cannot decrease
At a particular time, object cannot have multiple velocities
NOTE: Problem solving through graph is much easier and faster than theconventional method. So Practice solving problem through graph
8/6/2019 Physics Impt Point
4/34
Other Cases
Variable Acceleration:
The three equation of Motion is not applicablewhen acceleration varies.
The equation to be used for variable accelerationto be used are: V = ds /dt, a = dv/dt and a = v.dv/ds
Practice the concept of definite integralsincluding identifying the upper and lower limitsfor calculating the integral.
Relation between velocity (conditional cases):
Solve cases of Pulley Case of Ladder
Case of Wedge and block
Case of Wedge and Pulley
8/6/2019 Physics Impt Point
5/34
Relative Velocity
Measurement with respect to earth
Inertial frame of reference : references moving with uniform velocity
Relative Velocity Two dimension :Use vector for solving it.
Relative velocity Vs Difference of Velocity Relative velocity pertains to two objects. The order is also important
Difference of velocity can be for same or different object. Order is not important
Relative velocity Vs Resultant Velocity In relative velocity, two objects are separated
In resultant velocity, one object moves right within the second object. They are in contact with
each other. For eg boat moving in a river, wind flowing along with rain Relative velocity is calculated when the speed of both the objects are w.r.t same frame. The
resultant velocity is calculated for a particular object in a differenr frame (one w.r.t water andother w.r.t ground)
Relative velocity of an object w.r.t medium in which it is flowing = Velocity ofobject in still medium. The inherent ability of the object doesnt depend on whether medium is still or moving
Impt Points in River B
oat Problem: Shortest time is when the velocity of boat is perpendicular to the velocity of river Shortest path is when the boat reaches the opposite point Here the direction of boat is as
observed by the observer standing at the bank. This is not always possible For cases where velocity of boat w.r.t stream is greater than the velocity of stream, it is possible to
reach the directly opposite path
For cases where velocity of boat w.r.t stream is less than or equal to the velocity of stream, then theshortest path is when the drift is minimum. The conditions need to be calculated by differentiating.
8/6/2019 Physics Impt Point
6/34
Numericals on River Boat (http://cnx.org/content/m14034/latest/)
Example 1: A person can swim at 1m/s in still water. He swims to cross a
river of width 200m to a point exactly opposite to his initial position. If thewater stream in river flows at 2m/s in linear direction, then find the timetaken in secs to reach the opposite point
Example 2: A person can swim at a speed u in still water. He pointsacross the direction of water stream to cross a river. The water streamflows with a speed v in a linear direction. Find the direction in which heactually swims with respect to the direction of stream.
Example 3: A boat capable of sailing at 2m/s moves upstream in a river.The water stream flows at 1m/s. A person walks from the front end to therear end of the boat at a speed of 1m/s along the linear direction. What isthe speed of the person with respect to the ground.
Example 4: A boy swims to reach a point Q on the opposite bank, suchthat the line joining initial and final position makes and angle of 45 with
the direction perpendicular to the stream of water. If the velocity of waterstream is u, then find the minimum speed with which the boy shouldswim to reach his target.
Example 5: A boat crosses a river in minimum time, taking 10 mts duringwhich time it drifts by 120 m in the direction of stream. On the otherhand, boat takes 12.5 mts while moving across the river. Find (i) width ofriver (ii) velocity of boat in still water and (iii) speed of stream
8/6/2019 Physics Impt Point
7/34
Hints on Numericals on River Boat
Example 1: velocity of person relative to water = 1m/s. Theresultant velocity (velocity of person w.r.t ground) is along the yaxis. Ans: Not possible
Example 2: velocity of person relative to water = u . This directionis always perpendicular to direction of stream. Resultant velocity(velocity of person w.r.t ground) is at and angle theta with the x axis
where Tan(theta) = u/v Example 3: velocity of boat with respect to river = 2m/s (upstream).
Ans.. = 0m/s
Example 4: A boy swims to reach a point Q on the opposite bank,such that the line joining initial and final position makes and angleof 45 with the direction perpendicular to the stream of water. If thevelocity of water stream is u, then find the minimum speed withwhich the boy should swim to reach his target.
Example 5: Minimum time when velocity of person w.r.t river isperpendicular to velocity of river. Shortest path is when the startingand ending point are directly opposite. Ans..width = 200m, speedof stream = 12m/mt, velocity of boat w.r.t stream = 20 m /mt
Note: X axis is the flow of stream and y axis is perpendicular to X axis
8/6/2019 Physics Impt Point
8/34
Projectile
Following Cases to be seen:
Default case: Starting and ending point is at thesame vertical level
Important points
Relative change in height, range when one of the
variable say velocity or angle of projection isvaried. (Use concept of differentiation).
Inclined Plane:
Ending point is at the higher vertical level
Ending point is at the lower vertical level
Relative motion of two projectiles
Collision of projectiles
8/6/2019 Physics Impt Point
9/34
Important Results (Starting and Ending Point at the same Level)
)/1(
)2/(
2//2
/2
222
22
2
RxxTany
vSecgxxTany
gSinvhgSinvR
gvSinT
!
!
!
!
!
U
UU
U
U
U
Here:
V is the velocity of object projected
is the angle of projection with horizontal
T is the time of flight
R is range (horizontal distance)
h is the height (vertical distance)
Other Results:
Motion along perpendicular direction
Are independent of each other.
Range is maxm when = 45
Range is same for and (90 )
If h1 and h2 is the height for the
same range R then:
If T1 and T2 is the time of flight for
the same range R, then
If and are the angle made to anypoint from the two extremities of a
range in a projectile path of an object
projected with an angle from the
horizontal then
2*14 hh!
U2../. TanEKEP !
At highest Point:
g/*22*1 !
))(/( xxyananan !! UFE
8/6/2019 Physics Impt Point
10/34
Relative Change in the T, R, h when one of the independent variable is changed
)2(*)2/(sin/
)2(*)/2(sin/
)2(*)/(sin/
2
vgdvdh
vgdvdR
gdvdT
U
U
U
!
!
!
When is constant. The v is varied
)cos*sin*2(*)2/(/
)2(*)2(cos*)/(/
)(cos*)/2(/
2
2
UUU
UU
UU
gvddh
gvddR
gvddT
!
!
!
When v is constant. The is varied
RdRhdh // !
TdThdh /*2/ !
8/6/2019 Physics Impt Point
11/34
Important Results (Ending Point at the higher level)
)cos2/()(sin
)cos/()sin(*cos2
)cos/()sin(2
22
2
EEU
EEUU
EEU
gvh
gvR
gvT
!
!
!
Technique:
Resolve the motion along the incline and perpendicular to it
Resolve the component of g along the incline and perpendicular to it Solve independently for two directions.
Let:
V is the velocity of object projected
is the angle of projection with horizontal
is the angle of incline with the horizontalT is the time of flight
R is range (horizontal distance)
h is the height (vertical distance)
902 ! EURange is maximum when
8/6/2019 Physics Impt Point
12/34
Circular Motion
Horizontal: Here the weight (mg), which is pointing vertically downward is always
perpendicular to motion and as such doesnt contribute to centripetal
acceleration.
The centripetal acceleration is contributed by the Tension.
The vertical component balances the weight
The horizontal component (radially towards the circle) providecentripetal acceleration
The speed remains constant of an object.
Vertical circle: In this we use the concept of law of conservation of energy.
The body undergoes change in K.E. and P.E. The speed keeps on
changing
The weight mg along with tension T contributes towards centripetal
acceleration.
8/6/2019 Physics Impt Point
13/34
Types of Forces and its Direction
Gravitational Force:
This force always act vertically downward (along the ve y axis)
The force is assumed to be acting on the centre of Mass Reaction / Normal Force:
This force is exerted by the surface on which the body is in contact, away from the gravity
The direction is perpendicular to the tangent to the point of contact of the body with the surface.
Friction Force:
This force is perpendicular to the direction of the Normal force.
It is always away from the motion
Tension:
The force exerted by the end of a taut cord, string, or wire connected to a body is called the Tension
The direction of the tension-force is exerted along the direction of the cord
The magnitude of the tension is equal to the force that one would measure if the cord were cut and
a force spring-scale were inserted.
It is normally assumed that the cord is both massless and stretchless. When chord is massless the
tension is same across the chord. When the chord is inextensible, the acceleration is assumed to be
uniform across the string and the two bodies to which the string is connected.
When cord is both massless and stretchless, the forces at both ends of a cord have the same in
magnitude even if the cord changes direction through a pulley
At any moment the velocity and acceleration of two moving objects will be the same provided they
are connected by a massless, stretchless, taut-cord
8/6/2019 Physics Impt Point
14/34
Example on force and String
Different Forces:F1 is the Gravitational force = mg,
F2 is the reaction force and
F3 is the frictional force (if the body is sliding
downward) = F2 ( is the coefficient of friction)
Force resolution:The component of F1 along the y axis = -F1cos
(it is negative since the force is acting in the ve y axis)
The component of F1 along the x axis is = -F1sin
(It is negative since the force is along the ve x axis)
If this line
Doesnt cross
Through base,
The object
Will topple
8/6/2019 Physics Impt Point
15/34
Force Analysis in an accelerated frame of reference
Inertial frame (Ground) Accelerated frame (Lift, train)Case Description
A pendulum is suspended
from the roof of a train
compartment, which ismoving with a constant
acceleration a.
Find the deflection of the
pendulum bob from the
vertical as observed
from the ground and the
compartment
we consider a block of
mass 10 kg lying on the
floor of a lift, which itself is
moving up with an
acceleration of 2 m/s2.
Let g = 10 m/s2.
Find the weight measuredBy the spring balance .
8/6/2019 Physics Impt Point
16/34
Force Analysis in an accelerated frame of reference (Summary) Application of force analysis in accelerated frame of reference may have two approaches. We
can analyze using Newtons law from an inertial frame of reference. Alternatively, we can use
the technique of pseudo force and apply Newtons law right in the accelerated frame of
reference as described above. There is a school of thought that simply denies merit in pseudo force technique. The
argument is that pseudo force technique is arbitrary without any fundamental basis. Further,
this is like a short cut that conceals the true interaction of forces with body under
examination.
On the other hand, there are complicated situation where inertial frame approach may turn
out to be difficult to work with
Case Description: Here, a wedge isplaced on the
smooth surface of an
accelerated lift. We have to
study the motion of the block
on the smooth incline surface
of the wedge.Multiple accelerations here
complicates the situation.
The lift is accelerated with
respect to ground; wedge is
accelerated with respect to lift
(as the surface of the lift is smooth) ;
and finally block is accelerated with
respect to wedge (as wedge surface isalso smooth
Solution: In this case, it would bedifficult to assess or determine attributes
of motion by analyzing force in the
inertial ground reference. In situation
like this, analysis of forces in the non-
inertial frame of reference of the lift
eliminates one of the accelerations
involvedNote: when we analyze motion with
respect inertial frame of reference, then
all measurements are done with respect
to inertial frame. On the other hand, if
we analyze with respect to accelerated
frame of reference using concept of
pseudo force, then all measurements
are done with respect to acceleratedframe of reference
8/6/2019 Physics Impt Point
17/34
Important Cases to be seen
Accelerated frame of reference: For drawing F.B.D
When object is inside a frame which itself is accelerating: The force =ma in the opposite side of acceleration is added for newtons law to be
valid
An accelerated object moving in a circular path will have tangential
acceleration as well as centripetal acceleration, both at right angle to each
other. The resultant acceleration is the resultant of both.
An inclined object will topple before it slide if:
Angle is such that the centre of gravity at that instant no longer crosses the base of the
object
The angle is less than the limiting angle required as per friction.
Free Body Diagram: Practice for various cases
Sliding and Toppling One object over another:
Force on one body
Force on other body: When the object will move together. When they will move relative
to each other.
8/6/2019 Physics Impt Point
18/34
Work, Power and Energy
Work is done when Force acts over a displacement (The component of displacement is along
the direction of force, in same or opposite direction)
Work = F.d. It is a scalar quantity. Here F is the force, d is the displacement Work = F.d.Cos
Work is zero when: F = 0, d = 0 or Force is perpendicular to d
Gravitational Potential Energy: An object is said to have a potential energy if it has
the ability to do work by way of its position or state.
If an object of mass m is lifted to a height h, the change in P.E is given by:Ep = mgh As the object moves upward there is a increase in P.E due to work done by the external
force against gravity
The change in gravitational P.E depends upon change in vertical height. In an
inclined plane, change in P.E. depends upon change in vertical height.
Hookes Law: The stretch produced by a force applied to a string is directly
proportional to force applied, within elastic limit
Elastic Potential energy:
Work Energy Theorem:
In all cases, work done by a non-zero net force results in a change in K.E. but the
applied forces on an object may cause change in P.E. or K.E. or both.
2*2
1E kxp !
Wpk !(( EE
8/6/2019 Physics Impt Point
19/34
Isolated and Non-Isolated System
Isolated System: within an isolated system, energy may be transferred from one object to
another or transformed from one form to another, but it cannot be increased or decreased.
This is the law of conservation of energy. So Total Mechanical energy remains constant
Alternatively:
Conservative Forces: Forces that act within systems but do not change their
mechanical energy .Eg: Force of a gravity, elastic force,
Since the conservative force does not affect the mechanical energy of a system , the work done by a
conservative force to move an object from one point to another within the system is independent of
the path the object follows.
Non Conservative Forces: It causes mechanical energy of the system to change.
Eg friction force
Since the Work done by the friction force is negative, the final mechanical
energy is less than the initial mechanical energy. So, frictional force
reduces the mechanical energy of the system.
Comparision of Energy Position graph for Isolated and Non Isolated System: The component of the force of gravity parallel to the motion can be determined by calculating the
slope of the gravitational P.E. position graph
The net force can be determined by calculating the slope of the K.E. position graph
In the non-isolated system, the slope of the total energy curve gives the force of friction.
Power = Rate of doing work =
fm
fm
W
W
!
!
m12
m12
-
pk (!(
aveFVt
dF
t
WP !
(
(!
(!
8/6/2019 Physics Impt Point
20/34
More on Work and Collision
Work done when force is varying with distance
Area of F-X curve
Here F is the force, X is the displacement
Potential Energy: The notion of P.E. is applicable only to the class of forces where
work done against the force gets stored up as energy. When external constraints
are removed , it manifests itself as kinetic energy
The P.E. V(x) is defined if F(x) can be written as
This implies that
T
Collisions: Elastic: Both Kinetic Energy and the Linear momentum is conserved
Inelastic : Only Linear Momentum is conserved
Completely inelastic: After collision, both objects move together.
fi
f
VVdVdxx
V
Vi
xf
xi!! )(
!
f
i
x
x
fdxW
dx
dVx !)(
8/6/2019 Physics Impt Point
21/34
System of Particles and Rotational MotionParticle (point mass, no size) body of finite size (system of particles) Centre of Mass
Rigid body:
Definition: Perfectly definite and unchanging shape: No real body is truly rigid since real bodies deform under the influence of force.
Types of Motion:
Only Translational: All particles have same liner velocity at any instant of time
Only Rotational: All particles have same angular velocity at any instant of time
Translational + Rotational: Eg Rolling cylinder
Rotational Motion: Axis of Rotation: about which the body rotates
In rotation of rigid body about a fixed axis, every particle of the body moves in a circle
which lie in a plane perpendicular to axis of rotation and has its centre on the axis of
rotation.
For any particle on axis, the particle remains stationary while the body rotates
There are cases where axis may not be fixed only one of the point is fixed. Eg spinning
top whose nail is at a particular point (Here we assumed that the top is not slipping)
while its axis rotates to form a cone.
The movement of axis of the top around vertical is termed precession
Summary:
Motion of rigid body which is not pivoted or fixed may have either pure translational or
a combination of translation and rotation
Motion of a rigid body which is pivoted or fixed in some way is rotational
8/6/2019 Physics Impt Point
22/34
Some Important Formulae
)...mm)/(mxm...xmx(mX n21nn2211 !The Centre of Mass of a linear system:
The Centre of Mass in a 3 dimensional system:
)...mm)/(mrm...rmr(mR n21nn2211 !Here r is the position vector of the system of particles
R and r are both vector quantity.
Note: If the origin of the frame of reference is chosen to be centre of mass, then R = 0
For cases: where rigid body is considered as a continous distribution of mass (when
spacing between particle is small), we subdivide body into n small elements of mass
)m...,m,m n21 (((
If we make n bigger and bigger and each mass element
Smaller and smaller, we denote the sums by integrals
! xdmM *1
X ! ydmM *1
Y ! zdmM *1
Z
! rdm
M *
1
R
The Centre of Mass
Calculation
For a continous
Distribution of mass
8/6/2019 Physics Impt Point
23/34
Motion of Centre of Mass
!
!
!
!
!
!
!
!!
ext
ext
ii
FA
FFA
A
A
V
rmR
M
M
F...FFM
am...amamMdt
dvm...
dt
dvm
dt
dvm
dt
dVM
vm...vmvmM
dtdrm...
dtdrm
dtdrm
dtdRM
)rm...rmr(mM
int
n21
nn2211
n
n
2
2
1
1
nn2211
n
n
2
2
1
1
nn2211
The total internal force
Cancel each other due toNetwtons third law of motion
The Centre of Mass of a system of particles moves as if all the mass of a system
Was concentrated at the centre of mass and all the external forces were applied
At that point.
Further, to determine the motion of Centre of Mass, no knowledge of internal
force is required. Internal force do not contribute to motion of C.O.M.
8/6/2019 Physics Impt Point
24/34
ElectroStatics It deals with the study of forces, fields and potential energy arising out of static charge
State of Charge:
Static: Equilibrium state. No flow of charge - current Motion: Lead to flow of Current
Basic Properties of Charge:
Additivity
Charge is conserved
Charge is quantized
Coulombs Law: Fundamental law. Proved by experiment. For Point Charge: When linear size of charged body is much smaller than the distance
between them then the body can be considered as point charge.
It indicates the force between two point charges
The force is directly proportional to product of two charges and inversely proportional to
square of distance between them
It is a conservative force. Force is a vector qunatity. Its direction is along the line joining the two charged body
Like charges repel each other and unlike charges attract each other.
Force due to Multiple Charge:
Calculate the force due to each combination of charge using Couloms law. Magnitude
and direction
Calculate the resultant force using law of addition of vector
8/6/2019 Physics Impt Point
25/34
Electric Field Electric Field: due to a point charge at any point is defined as a force
experienced by a unit charge at that point.
It is a vector quantity. It is emnated in all directions (3 Dimensional) For +charge, it is directed radially outward
For ve charge, it is directed radially inward.
Electric field is independent of test charge.
Here test charge is taken to be near to zero to reduce the interference effect that the
test charge may have on the existing field of Source Charge.
It can be concluded that Electric field is same for points which are at a same distancefrom the Source Charge.
Thus electric field due to a point charge at the centre of sphere is same across all the
points of sphere. It has a spherical symmetry
Electric field due to a system of Charges:
Calculate the electric field for each point charge at a particular point
It will have both magnitude and direction
The resultant electric field can be calculated using vector addition.
Force experienced by a Charge q in Electric Field
The magnitude is given by F = qE
The direction is given by: Along the Electric field for positive charge
Opposite to electric field for negative charge
)(0
Eq
F
q
Lim
p
!
8/6/2019 Physics Impt Point
26/34
Electric Field Lines and Electric Flux Electric Field Lines:
For representing Electric Field
Its is denser where the electric field is strong. It is the relative density across differentpoints that matter.
The tangent to the electric field lines gives the direction of electric field at that point
Two Electric Field lines never intersect each other
Electrostatic field lines do not form a closed loop. This follows from the conservative
nature of electric field.
A field line is a space curve: That is curve in three dimensions In a charge free region, electric field lines can be taken to be continous curves without
any break.
Electric Flux:
The number of field lines crossing an area placed normal to Electric field
If is the angle between the Electric field and the normal to area dA and assuming that
electric field is constant in that area then , then Electric flux
= E.dA = EdA Cos
The direction of dA is a normal vector from area element dA and pointing outward of a
closed surface,
The electric flux is zero if Electric Field is zero
Electric field is parallel to the normal of the area
8/6/2019 Physics Impt Point
27/34
Electric Dipole Pair of equal and opposite point charge separated by a distance 2a.
Dipole Moment : Denoted by p.
It is vector quantity. Direction is along the line joining two point charges from negativeto positive
Magnitude is given by: 2aq
Electric field at a point along axis:
At a distance r from the centre of dipole
Here p is a unit vector
For distance (r>>a)
Electric field at a point along the equatorial plane:
At a distance r from the centre of dipole
For distance (r >>a)
Impt Point:
Dipole field at a large distance falls of not as 1/r2 but as 1/r3
The eqn (for r>>a) is exact for any r for point dipole (size 2a approaches zero the charge
q approaches infinity in such a way that 2aq is finite)
par
arq222 )(4
4*E
!
TI
pr
aq3
4
4*
TI!
par
aq2/322 )(4
2*E
!
TI
pr
aq3
4
2*E
TI!
8/6/2019 Physics Impt Point
28/34
Electric Dipole Contd Physical Significance of Electric Dipole:
In most molecules, the centre of positive and negative charge lie at the same place. So
their dipole moment is Zero. They are called non-polar molecule. Eg CO2, CH4. Theydevelop a dipole moment when electric field is applied.
In some molecules, centre of positive and negative charge do not coincide. Thus they
have a permanent dipole moment even in the absence of electric field. Eg water
Dipole in a Uniform External Electric Field:
The net force is Zero
The torque (Rotational force) is given by = PXE = 2aqSinp The direction of torque will be such that it will tend to align the dipole in the direction of
electric field.
Dipole in a Non-Uniform External Electric Field:
The net force is not zero
Torue will also be there
So dipole will be subjected to both linear and rotational motion
8/6/2019 Physics Impt Point
29/34
Continous Charge Distribution Continous charge Distribution: Charge is distributed uniformly across the object
Ring: Eg bangle. (negligible thickness). Circumference = 2r. Linear Charge density
Disc: We use Area: We use Surface Charge density = Q / Area of object Sphere: We use Volume charge density = Q / Volume of Object
For Calculating Electric Field in a Continous Charge:
Take a small area where charge is constant
Calculate the Electric field due to that small segment of Charge
The total electric field is sum of electric field due to such small segment across the source body
For linear Object: Wire or Ring
Linear charge density = Q/l or Q/2r.
For Surface charge density: Eg disk
A disk is considered to be made up of muliple ring of varying radii of negligible thickness joined
together
We take a ring at a radius r of thickness dr Its Area is given by 2r*dr
For Volume Charge density: Eg Solid Sphere
Made up of multiple disk of varying radii of negligible thickness and joined together
We take a test sphere of radius r and thickness dr.
Its Volume is given by = 4r2 * dr.
8/6/2019 Physics Impt Point
30/34
Gausss Law Used for Calculating Electric Field
Definition: Total Electric Flux through a closed surface = Net Charge Enclosed / Epsilon
In the situation, when the surface is so chosen that are some charges inside and some outside, the
electric field (whose flux appears on the L.H.S.,) is due to all charges, both inside and outside. The
term q on the R.H.S. is however the net charge enclosed by the surface.
If the total electric flux through a closed surface is zero then the net charge enclosed by the surface is
zero.
Gausss Law is true for any closed surface no matter what is its shape or size.
Gaussian Surface: A Closed surface . Ensure that Gaussian surface doesnt cross through any discrete
charge as electric field is not well defined at the location of the charge.
For continous volume charge distribution, it is defined at any point in the distribution
For surface charge distribution, electric field is discontinous across the surface.
With Gaussian symmetry such that Electric field is constant in magnitude at that point. All points are
equivalent to given charge distribution. Typical Gaussian Surface is Sphere, Cylinder
TheGausss Law is based on the inverse square dependence on distance contained in the Coulombs
law. Any violation ofGauss Law will indicate departure from the inverse square law.
Note:
The electric field due to a charge configuration with total charge zero is not zero; but for
distances large compared to the size of the configuration, its field falls of faster than 1/r2, typical
of field due to a point charge. An electric dipole is the simplest example of this.
8/6/2019 Physics Impt Point
31/34
Application ofGausss Law (Do Yourself)
Electric field due to a infinite long straight uniformly charged wire
Electric field due to a uniformly charged infinite plane Sheet
Electric Field due to a uniformly charged thin spherical shell
Electric field due to a uniformly charged insulating sphere
8/6/2019 Physics Impt Point
32/34
Conductors
All the Charge in the conductors lies on the external surface
The electric field is zero inside the conductor
If the conductor is enclosing a insulated cavity with charge q then
Charge q will be induced on the inside of the conductor. This is because for electric field
inside the surface of conductor to be zero, the net charge shall be zero. This will be zero
only when the charge q which is in the cavity is nullified by the equal and opposite
charge. This charge has to be at the inside of the surface of the conductor
Chare +q will be there on the outside of the conductor to compensate for the charge q
on inside of the surface.
Q1
Q2
Q3
Initial State
Q1
Final State
-Q1
8/6/2019 Physics Impt Point
33/34
Electrostatic Potential Energy
Potential Energy: When an external force does work in moving a body from one point to
another against a force such as gravitational force or spring force of electrostatic force, that
force gets stored as potential energy in the object. When the external force is removed, the body gains kinetic energy and loses potential energy
The sum of kinetic and potential energy is thus conserved.
Such forces are called conservative forces
Electrostatic Potential Energy: Work done by an external force in moving a test charge q from
point R to point P, against that force. Two points to be taken care are:
The test charge q is so small that it has no effect on the source Charge The body is moved from point R to P with zero acceleration i.e at any instant the net
force acting on the charged body is zero. This is possible when
Thus Work done by external force
This work is stored as Potential Energy = U = Up Ur = Wrp Impt. Points regarding Electrostatic Potential Energy:
Depends only on the initial and final position and not on the path followed. This is a
fundamental characteristic of conservative force
It is the difference of potential energy that is significant and not the absolute potential
energy. So there is freedom in choosing a point where potential energy is zero.
A convenient choice is to take potential energy as zero at infinity
Eext F!F
drFdrFW
P
R
P
R
extERP .. !!
8/6/2019 Physics Impt Point
34/34
Electrostatic Potential
Electrostatic Potential: due to a Source Charge Q At a point is defined as work done in
bringing a unit positive charge q from infinity to that point. It is denoted by V.
It is a scalar quantity. Potential due to a point Source charge Qat a distance r
Thus E at a point is inversely proportional to square of distance while V is inversely
proportional to distance. Thus E falls off faster than potential
Potential Due to Electric dipole
For distance r >>a then
Here is the angle between p and radius vector joining the mid point of dipole to point The above eqn is approx. equal
For point dipole: The said equation is exact
For point along the axis, = 0 or 180, so
For point along the equatorial line, = 90 so V = 0
Potential due to System of Charge: Since potential is a scalar quantity, the potential due to a
system of charge is the algebraic sum of potential due to each charge at a point,
The potential inside the conducting shell is constant
Equipotential Surface: The surface with the same potential at all points
Any point on a sphere of radius r enclosing a point charge q is at equipotential
No work is done in moving a body from one point to other in a equipotential surface
The electric field line is normal to equipotential surface
r
Q
TI4V !
24
**2V
r
Cosqa
TIU
!
24
*2V
r
qa
TI!