Potential Difference Potential Field Potential Gradient...

Indraprastha Institute of

Information Technology Delhi ECE230

Lecture – 12 Date: 03.02.2014 • Potential Difference • Potential Field • Potential Gradient • Electric Dipole



• Say A and B are two points somewhere on a circuit. But let’s call these points something different, say point + and point - .

(r).C

V E dl + -

• V represents the potential difference (in volts) between point (i.e., node) + and point (node) - . Note this value can be either positive or negative.

Q: But, does this mean that circuits produce electric fields?

Potential Difference



A: Absolutely! Anytime you can measure a voltage (i.e., a potential difference) between two points, an

electric field must be present!

Potential Difference (contd.)



Potential Difference (contd.)

• Note that VAB can be expressed as:

(r). (r ) (r )AB A B

C

V E dl V V

where point A lies at the beginning of contour C, and B

lies at the end.

• The potential at any point is the potential difference between that point and a chosen point (or reference point) at which the potential is zero (usually ground !).

( ).

r

V E r dl



Polarity of Electric Potential and Fields are Opposite?

𝐸 𝐸 𝐸 𝐸

• Lets consider the simple case of a positive charge 𝑞 in a uniform electric

field 𝐸 = −𝐸𝑎 𝑦

• The presence of 𝐸 exerts a force

𝐹 𝑒 = 𝑞𝐸 on the charge in the –y direction

𝐹 𝑒

• To move the charge along the +y direction against 𝐹 𝑒, there is a need of

external force 𝐹 𝑒𝑥𝑡 to counteract 𝐹 𝑒 • To move 𝑞 without acceleration:

𝐹 𝑒𝑥𝑡

ext eF F qE

• The work done in moving the charge a vector differential distance dy = 𝑑𝑦𝑎 𝑦:

𝑑𝑦

ˆ. .ext ydW F dya qE dy



Polarity of Electric Potential and Fields are Opposite?

• Now, the electric potential is the energy per unit charge:

dWdV

q .dV E dy

Conclusion of our Premise !



Electric Potential for Point Charge

• Recall that a point charge Q, located at the origin ((𝑟′ = 0), produces a static electric field:

• Now, we know that this field is the gradient of some scalar field:

Q: What is the electric potential function V(𝑟 ) generated by a point charge Q, located at the origin?

Q: Where did this come from ? How do we know that this is the correct solution? A: We can show it is the correct solution by direct substitution!

2

0

ˆ(r)4

r

QE a

r

(r) ( )E V r

A: We find that it is: 0

(r)4

QV

r



(r) ( )E V r 04

Q

r

'

0

ˆ 0 04

r

Qa

r r

The correct result!

Q: What if the charge is not located at the origin ? A: Substitute r with |𝑟 − 𝑟′ |, and we get:

Electric Potential for Point Charge (contd.)

Verification:

2

0

ˆ(r) ( )4

r

QE V r a

r

0

(r)4 '

QV

r r

where, as before, the position vector 𝑟′ denotes the location of the charge Q, and the position vector 𝑟 denotes the location in space

where the electric potential function is evaluated.



• The scalar function V(𝑟 ) for a point charge can be shown graphically as a contour plot:





• Or, in three dimensions as: Note the electric potential increases as we get closer to the point charge (located at the origin). It appears that we have “mountain” of electric potential; an appropriate analogy, considering that the potential energy of a mass in the Earth’s gravitational field increases with altitude (i.e., height)!




• Recall the electric field produced by a point charge is a vector field that looks like:



• Combining the electric field plot with the electric potential plot, we get:


Given our understanding of the gradient, the above

plot makes perfect sense! Do you see why ?



Electric Potential Function for Charge Densities

• Recall the total static electric field produced by 2 different charges (or charge densities) is just the vector sum of the fields produced by each:

• Since the fields are conservative, we can write this as:

1 2(r) (r) (r)E E E

1 2(r) (r) (r)E E E 1 2( ) ( ) ( )V r V r V r

1 2( ) ( ) ( )V r V r V r

• Therefore, we find: 1 2( ) ( ) ( )V r V r V r

In other words, superposition also holds for the electric potential function! The total electric potential field produced by a collection of charges is

simply the sum of the electric potential produced by each.



• Consider now some distribution of charge, ρ𝑣(𝑟 ). The amount of charge

dQ, contained within small volume dv, located at position 𝑟′ , is:

• The electric potential function produced by this charge is therefore:

• Therefore, integrating across all the charge in some volume v, we get:

Electric Potential Function for Charge Densities (contd.)

( ') 'vdQ r dv

0 0

( ') '( )

4 ' 4 '

vdQ r dvdV r

r r r r

0

( ')( ) '

4 '

v

v

rV r dv

r r



• Likewise, for surface or line charge density:


0

( ')( ) '

4 '

S

S

rV r dS

r r

0

( ')( ) '

4 '

l

C

rV r dl

r r

Note that these integrations are scalar integrations—typically they are easier to evaluate than the integrations resulting from

Coulomb’s Law.

• Once we find the electric potential function V 𝑟 , we can then determine the total electric field by taking the gradient:

(r) ( )E V r



• Thus, we now have three (!) potential methods for determining the electric field produced by some charge distribution ρ𝑣(𝑟 ).

1. Determine 𝐸 𝑟 from Coulomb’s Law.

2. If ρ𝑣(𝑟 ) is symmetric, determine 𝐸 𝑟 from Gauss’s Law. 3. Determine the electric potential function V 𝑟 , and then determine

the electric field as 𝐸 𝑟 = −∇V(𝑟 ).

Q: Yikes! Which of the three should we use?? A: To a certain extent, it does not matter! All three will provide the same result (although ρ𝑣(𝑟 ) must be symmetric to use method 2!).


However, if the charge density is symmetric, we will find that using Gauss’s Law (method 2) will typically result in much less work!

Otherwise (i.e., for non-symmetric ρ𝑣(𝑟 )), we find that sometimes method 1 is easiest, but in other cases method 3 is a bit less stressful

(i.e., you decide!).



Example – 1

• Determine the electric potential at the origin due to four 20-mC charges residing in free space at the corners of a 2𝑚 × 2𝑚 square centered about the origin in the x–y plane.

• For four identical charges all equidistant from the origin:

' 2R r r m

6 5

00

4 20 10 2 10( )

4 2V r

(V)

0

4( )

4

QV r

R



Example – 2

• A spherical shell of radius R has a uniform surface charge density ρS. Determine the electric potential at the center of the shell.

2

0 0

( ) 44

S S RV r R

R

0

( ')( ) '

4 '

S

S

rV r dS

r r

0

'( )

4 '

S

S

dSV r

r r

0

( ) '4

S

S

V r dSR



Relationship between 𝑬 and V

• We have learnt that the electrostatic field is conservative and therefore, following is true for the given situation:

BA ABV V . 0BA AB

L

V V E dl

Line integral of 𝐸along a closed path is zero

• Lets apply Stoke’s theorem: . 0 .L S

E dl E dS 0E

Conservative or Irrotational



Relationship between 𝑬 and V (contd.)

0E . 0BA AB

L

V V E dl

• We defined potential as: .V E dl .dV E dl

x y zdV E dx E dy E dz

• Alternatively we can also write: V V VdV dx dy dz

x y z

• Comparison gives: x

VE

x

y

VE

y

z

VE

z

• Therefore: E V

Maxwell’s equations for Electrostatics Intergal Form

Differential Form



Relationship between 𝑬 and V (contd.)

E V

The electric field intensity is the gradient of 𝑉. The

negative sign shows that the direction of 𝐸 is opposite to

the direction in which V increases ↔ 𝐸 is directed from higher to lower levels of V

It provides another tool to determine electric field apart from Coulomb’s

and Gauss’s laws → 𝐸 can be obtained if the scalar function V is known



Example – 3

• Determine Electric Field due to potential: 𝑉 = ρ2 𝑧 + 1 𝑠𝑖𝑛ϕ

E V 1

ˆ ˆ ˆz

V V VE a a a

z

2ˆ ˆ ˆ2 ( 1)sin ( 1)cos sin zE z a z a a

2ˆ ˆ ˆ2 ( 1)sin ( 1)cos sin zE z a z a a



Example – 4

• Determine Electric Field due to potential: 𝑉 = 𝑒−𝑟𝑠𝑖𝑛θ𝑐𝑜𝑠2ϕ

E V 1 1

ˆ ˆ ˆsin

r

V V VE a a a

r r r

1ˆ ˆ ˆsin cos2 cos cos2 ( 2sin 2 )

rr r

r

eE e a e a a

r r

1 2ˆ ˆ ˆsin cos2 cos cos2 sin 2

rr r

r

eE e a e a a

r r



Poisson’s and Laplace’s Equation

• From Gauss’s Law: 0

. vE

E V • We have:

0

. vV

2 2 22

2 2 2

0

vV V VV

x y z

(Poisson’s Equation)

• If the medium under consideration contains no charge then: 2 0V

Laplace’s Equations

These formulations are extremely useful for determining the electrostatic potential V in regions with boundaries on which V is known, such as the

regions between the plates of a capacitor with specified voltage difference across it.



Electric Dipole

• An electric dipole is formed when two point charges of equal magnitude but opposite signs are separated by a small distance → a pretty useful configuration!!!

• In practical situation, the distance from the point of interest is much greater than the separation.

• The potential at point 𝑃(𝑟, θ, ϕ) is: P

+Q

-Q

r1

rr2

cos

2

d

za

d

• Since 𝑟 ≫ 𝑑; r, r1, and r2 are almost parallel.

2 1

0 1 2 0 1 2

1 1

4 4

Q Q r rV

r r r r

1 cos2

dr r 2 cos

2

dr r

2 1 cosr r d

2

1 2r r r• Furthermore:



Electric Dipole

2

0

cos

4

Q dV

r

• Since, dcosθ = 𝑑 . 𝑎𝑟 2

0

ˆ.

4

rQd aV

r

Qd = p is the dipole moment

Dipole moment is directed from –Q to +Q

• If the dipole center is not at the origin but at r’ then:

• The electric field due to the dipole with center at the origin:

1ˆ ˆ

r

V VE V a a

r r

2

0

ˆ.

4

rp aV

r

3

0

. '

4 '

p r rV

r r

3 3

0 0

cos sˆ ˆ

2 4r

Qd Qd inE a a

r r



3

0

ˆ ˆ2cos s4

r

QdE a in a

r

• It is important to notice that a point charge is a monopole and its 𝐸 varies inversely as r2 while its V varies inversely as r. For a dipole, the respective

variations are 𝐸 ∝1

r3 and V∝

1

r2 while 𝐸 due to successive higher-order

multipoles vary inversely as r4, r5, r6, …., and their corresponding V vary inversely as r3, r4, r5, ……

Electric Dipole

3

0

ˆ ˆ2cos s4

r

pE a in a

r

p p Qd



Example – 5 • Point charges 𝑄 and −𝑄 are located at (0,

𝑑

2, 0) and 0,−

𝑑

2, 0 . Show that

at point 𝑟, θ, ϕ , where 𝑟 ≫ 𝑑,

2

0

sin sin

4

QdV

r

• Find the corresponding 𝐸 as well.

2

0

ˆ.

4

rp aV

r ˆ ˆ ˆ. . sin sinr y rp a Qda a Qd

2

0

sin sin

4

QdV

r

E V 1 1

ˆ ˆ ˆsin

r

V V VE a a a

r r r

3 3 3

0

2sin sin cos sin cosˆ ˆ ˆ

4r

QdE a a a

r r r

• The dipole is oriented along y-axis. Therefore:

Now:

3

0

ˆ ˆ ˆ2sin sin cos sin cos4

r

QdE a a a

r

Potential Difference Potential Field Potential Gradient...

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