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Jackson E&M Solution Homework 2, #2 .

This document discusses the electrostatic solution to a problem involving a conducting spherical shell with a point charge inside. 1) The method of images is used to find an expression for the potential involving the original point charge and an image charge. 2) The surface charge density on the inner surface of the shell is calculated and found to be a function of the angle. 3) The induced charge on the inner surface due to the interior point charge is calculated to be equal and opposite to the original point charge.

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Tikhon Bernstam
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2K views1 page

Jackson E&M Solution Homework 2, #2 .

This document discusses the electrostatic solution to a problem involving a conducting spherical shell with a point charge inside. 1) The method of images is used to find an expression for the potential involving the original point charge and an image charge. 2) The surface charge density on the inner surface of the shell is calculated and found to be a function of the angle. 3) The induced charge on the inner surface due to the interior point charge is calculated to be equal and opposite to the original point charge.

Uploaded by

Tikhon Bernstam
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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PHY 5346

HW Set 2 Solutions – Kimel

5. 2.2 The system is described by

a) Using the method of images


1 q q′
φx⃗ = +
4π 0 |x⃗ − ⃗
y| |x⃗ − ⃗y′ |
with y ′ = ay , and q ′ = −q ay
2

b) σ = − 0 ∂n∂ φ|x=a = + 0 ∂x∂ φ|x=a


1 ∂ q q′
σ = 0 +
4π 0 ∂x x 2 + y 2 − 2xy cos γ
1/2
x 2 + y ′2 − 2xy ′ cos γ
1/2

y2
a 1−
σ = −q 1 a2
4π y 2 + a 2 − 2ay cos γ 3/2
Note
y2
q induced = a 2 ∫ σdΩ = −q 1 a 2 2πa 1 − 2 ∫ −1
1
dx , where x = cos γ
4π 3/2
a y 2 + a 2 − 2ayx
q 2
q induced = − aa 2 − y 2  = −q
2 aa 2 − y 2 
c)
qq ′ 1 q 2 ay
|F| = = , the force is attractive, to the right.
4π 0 y ′ − y 2 4π 0 a 2 − y 2 
d) If the conductor were fixed at a different potential, or equivalently if extra charge were put on
the conductor, then the potential would be
q q′
φx⃗ = 1 + +V
4π 0 |x⃗ − ⃗y| |x⃗ − ⃗y′ |
and obviously the electric field in the sphere and induced charge on the inside of the sphere would
remain unchanged.

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