Prof. David R. JacksonECE Dept.

Fall 2014

Notes 15

ECE 2317 Applied Electricity and Magnetism

1

Potential Integral Formula

This is a method for calculating the potential function directly, without having to calculate the electric field first.

This is often the easiest way to find the potential function (especially when you don’t already have the electric field calculated). There are no vector calculations involved.

The method assumes that the potential is zero at infinity. (If this is not so, you must remember to add a constant to the solution.)

2

Potential Integral Formula (cont.)

0 04 4

v r dVdQd

R R

04

v

V

r dVr

R

Integrating, we obtain the following result:

0 x

y

zr (x, y, z)

R

v r r

3

From the point charge formula:

04

q

r

Point charge formula:

Potential Integral Formula (cont.)

Summary for all possible types of charge densities:

04

s

S

r dSr

R

04C

r dlr

R

04

v

V

r dVr

R

4

Note that the potential is zero at infinity (R )

in all cases.

Example

Find (0, 0, z)

0

0

0

2 2

20

2 20 0

0

2 20

4

4

24

C

dr

R

R z a

a dr

z a

a

z a

Circular ring of line charge

x

y

z

a

R

r = (0, 0, z)

l0 [C/m]

5

Note: The upper limit must be larger than the lower limit, to keep dl positive.

Example (cont.)

0

2 20

1V

2

ar

z a

z For

00

0 0

0

2

2 4

4

aar

z z

Q

z

(This agrees with the point charge formula.)

6

Example

7

Find (0, 0, z)

0

0

04v

V

dVr

R

Solid cube of uniform charge density

x

y

z

R

r = (0, 0, z)

v0 [C/m]

/2 /2 /20

22 20 /2 /2 /24

a a av

a a a

dx dy dzz

x y z z

2 2 2R x x y y z z

a

a

a

The integral can be evaluated numerically.

Example (cont.)

8

0 10 20 30 40 500

0.2

0.4

0.6

0.8

10.901

0.02

F h( )

501.1 hz [m]

[V]

0 04

1.0v

a m

Result from Mathcad

x

y

z

(0, 0, z)

a

a

a

Example (cont.)

9

Result from Mathcad

z [m]

0 04

1.0v

a m

1 1.5 20

0.5

1

1.5

21.767

0.5

F h( )

20.51 h0.5 1.0 1.5 2.0

0

0.5

1.0

1.5

2.0

[V]

x

y

z

(0, 0, z)

a

a

a

Face of cube

1.793 V

0.5 [m]

z

z

as

Limitation of Potential Integral Method

This method always works for a “bounded” charge density; that is, one that may be completely enclosed by a volume.

10

For a charge density that extends to infinity, the method might fail because it may not be possible to have zero volts at infinity.

This will happen when there is an infinite voltage drop going to infinity (e.g., infinite line charge problems, or 2D problems in general).

Example of Limitation (cont.)

0

0

0

2 20

4

1

4

l

l

dzR

dzz

Here the potential integral formula fails.

The integral does not converge!x

y

z

l0 [C/m]

r

R

z

11Infinite line charge

Example of Limitation (cont.)

12

Infinite line charge

x

y

z

l0 [C/m]

r

R ( = b)

b

0

0

ln V2

b

The field-integration method still works:

(From Notes 14)

Note: We can still use the potential integral method if we assume a

finite length of line charge first, and then after solving the problem let

the length tend to infinity.

(This will be a homework problem.)