Prof. D. Wilton ECE Dept. Notes 15 ECE 2317 Applied Electricity and Magnetism Notes prepared by the...

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Prof. D. WiltonECE Dept.

Notes 15

ECE 2317 ECE 2317 Applied Electricity and MagnetismApplied Electricity and Magnetism

Notes prepared by the EM group,

University of Houston.

Potential Integral FormulaPotential 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)

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.)

Potential Integral Formula (cont.)Potential Integral Formula (cont.)

x

y

z

v (r´)

r (x, y, z)

R

0

0 0

' '

4 4v r dVdQ

dR R

0

' ', 0

4v

V

r dVr

R

Integrating,we obtain the following result:

Potential Integral Formula (cont.)Potential Integral Formula (cont.)

Similarly

0

0

' '

4

' '

4

s

S

C

r dSr

R

r dlr

R

ExampleExample

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

x

y

z

a

R

r = (0, 0, z)

l0

Example (cont.)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

(agrees with the point charge formula)

Limitation of Potential Integral MethodLimitation of Potential Integral Method

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

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

Example of LimitationExample of Limitation

0

02E

Assume that () = 0

The integral does not exist!

0

0

0

0

0

0

0

1

2

ln2

r

r

r

r

E dr

E d z dz d

E d

d

infinite line charge

x

y

z

l0 [C/m]

r

Example of Limitation (cont.)Example of Limitation (cont.)

0

0

0

2 20

2 20

0

4

1

4

ln4

l

l

l

z

dzR

dzz

z z

If we try to use the potential integral formula anyway:

The integral does not converge!

x

y

z

l0 [C/m]

r

R

Sharp Point PropertySharp Point Property

A sharp point produces a strong electric field

b a

qbqa

Q [C]

Determine: How much charge goes to each sphere, and the electric field at the surface of each sphere.

The system is charged with Q [C].

0 04 4

a b

a bq q

a b

a bq q Q Also,

a

aq Q

a b

b

bq Q

a b

b a

qbqa

Q [C]

Solution:

a

b

q a

q b

Sharp Point Property (cont.)Sharp Point Property (cont.)

(More charge on the larger sphere.)

20

20

2

4

4

ara

brb

ra a

rb b

qE

a

qE

b

E qb

E a q

ErbEra

ra

rb

E b

E a

Hence, (A stronger electric field exists on the smaller sphere)

so

Sharp Point Property (cont.)Sharp Point Property (cont.)

V

sharp point

A high electric field is created at a sharp metal point.

+

-

Sharp Point Property (cont.)Sharp Point Property (cont.)

Lightning RodLightning Rod

- - - - - - - - - - - - - -

Lightning is “attracted” to the rod, not the house.

Important to have it grounded !