Prof. David R. JacksonECE Dept.

Spring 2016

Notes 14

ECE 3318 Applied Electricity and Magnetism

1

Potential Calculation from Field

2

In this set of notes we explore calculating the potential function (x,y,z) (assuming that we already know the electric field).

Note: The electric field is found first, either from Coulomb’s law (superposition) or from Gauss’s law.

Choose:

Potential Formula

B

ABA

V E dr

A B

R

r

r R E dr

r

R

r R E dr 3

or

A r B R

r

R

Then we have:

Potential Formula (cont.)

r

R

r R E dr

Cr

R

4

1) Pick a reference point (if it is not already given).2) Integrate in a coordinate system of your choice (remember that any path can be chosen in statics).

Recipe:

Path Independence of Voltage Drop

20

0

0

ˆ ˆˆ ˆ sin

4

4

1 14

B

A

B

A

B

A

B

AB rA

r

rr

r

r

r

r

A B

V r E r dr r d r d

E dr

q drr

qr

qr r

Consider first a point charge:

By superposition, the potential from any set of charges is path independent in statics.

The integral is path independent (the answer only depends on the endpoints).

5

y

z

q

Point charge

x

A

CB

Example

Given: R = , (R) = 0

20

ˆ4

qE rr

Choose a radial path for convenience.

Find the potential function

r

R

r R E dr 0r

R

E dr

y

z

q

r

RC

Point charge

x

6

Example (cont.)

20

0

ˆ ˆˆ ˆ sin

4

4

r

rR

r

rR

r

r

r E r dr r d r d

E dr

q drr

qr

0

V4

qrr

7

Hence, we have

Summary for a Point Charge

0

V , 04

qrr

8

20

V/mˆ4

qE r rr

y

z

q

r

x

rThese results for a point charge form the building blocks for all

other charge distributions (using superposition).

This is done in Notes 15.

ExampleGiven: R is at = b, (R) = 0

0

0

ˆ2

E

Find the potential function

Choose a radial path

0

ˆˆ ˆ ˆ

r

R

r

R

b

E dr

E d z dz d

E d

Infinite line charge

9

x

y

z

l0 [C/m]

r

R ( = b)

C

b

r

R

r R E dr

Example (cont.)

0

0

0

0

0

0

2

ln2

ln2

b

b

d

b

Note: b cannot be chosen as (there is an infinite voltage drop between = and = ).

0

0ln V

2b

10

In 2D problems (infinite in z) the reference point cannot be put at infinity.

Example

x

y

z

a

l0 [C/m]

r (0,0,z)

Given: () = 0

Note: We only know E on the z axis (from a previous example), so we MUST choose a path on the z axis.

Find the potential function on the z axis

Circular ring of line charge

11

r

R

r R E dr

Example (cont.)

() = 0

C

x

y

z

r

R ()

Choice of path:

12

r

R

r R E dr

Example (cont.)

From Coulomb’s law, we know that on the z axis the field is

ˆ ˆˆ ˆr

zR

z

z

r z E xdx y dy z dz

E dz

03/22 2

0

0, 0,2z

a zE zz a

13

0r

R

r E dr

r

R

r R E dr

so

Example (cont.)Hence,

03/22 2

0

03/22 2

0

1/22 20

0

0, 0,2

2

2

z

z

z

a zz dzz a

a z dzz a

a z a

02 2

0

10, 0, V2

azz a

The result is:

14

Adding a Constant to

Proof:

2

22 2

r

R

r R E dr

1

11 1

r

R

r R E dr

1

2

1

2

1 21 2 1 2

1 21 2

Rr

R r

R

R

r r R R E dr E dr

R R E dr

constant

C1

R1

r

C2

R2

15

1 1 01 2 2 02,R R

Assume two solutions:

Two different solutions for the potential function can only differ by a constant.

Adding a Constant to (cont.)

2 1, , , ,x y z x y z C

Conclusion:

Valid potential functions can only differ by the addition of a constant.

Adding a constant to a valid potential function gives a another valid potential function (this does not change the electric field).

16

ExampleGiven: (z = 1 [m]) = 10 [V]

Find the potential function on the z axis

02 2

0

10, 0,2

az Cz a

Start with our previous solution, which has zero volts at infinity, and add a constant to it:

Circular ring of line charge

x

y

z

al0 [C/m]

r (0, 0, z)

R (0, 0, 1)10 [V]

17

Example (cont.) 0

2 20

10, 0,2

az Cz a

0, 0,1 10 [V]

02 2

0

1 102 1

a Ca

Set:

x

y

z

a

l0 [C/m]

r (0, 0, z)

R (0,0,1)10 [V]

18

02 2

0

110 [V]2 1

aCa

Hence we have:

02 2 2 2

0

1 10, 0, 10 [V]2 1

azz a a

The solution is thus: