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## Homework Statement

(Not for homework/assignment. Just doing problems for practice)

This is from Griffiths Introduction to Electrodynamics, 4

^{th}edition, p.112 Problem 2.60

" A point charge

*q*is at the centre of an uncharged spherical conducting shell of inner radius

*a*and outer radius

*b. Question:*How much work would it take to move the charge out to infinity(through a tiny hole drilled in the shell)? [Answer: ##

(q^2/4\pi\epsilon_0)(1/a)

## ]

.

.

I later found out that the answer provided in the book is wrong (printing error) and the correct answer is ##\frac{q^2}{4\pi\epsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right)##

Source: http://www.reed.edu/physics/faculty/griffiths/ED4a.pdf

## Homework Equations

$$W=\frac{\epsilon_0}{2}∫E^2d\tau$$

$$V(r_1)-V(r_2)=\frac W q$$

$$W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i)$$

## The Attempt at a Solution

My first approach was to use the equation ##W=\frac{\epsilon_0}{2}∫E^2d\tau## (integrating all over space)

We have using Gauss's law,

$$\vec E=\begin{cases}

\frac{q}{4\pi\epsilon_0r^2} \hat r &r>b\\

0&a\leq r\leq b\\

\frac{q}{4\pi\epsilon_0r^2} \hat r&r<a

\end{cases}$$

Substituting, and using spherical coordinates, we have

$$\frac{q^2}{8\pi\epsilon_0}\left[ \int_0^a\frac{1}{r^2}dr+\int_b^∞\frac{1}{r^2}dr\right]$$

$$=\frac{q^2}{8\pi\epsilon_0}\left[\frac{-1}{a}+\frac{1}{0}-\frac{1}{∞}+\frac{1}{b}\right]$$

This leads to infinity.

.

.

Approach 2) Using the equation ##W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i)## , with our 3 'objects' being : the point charge at the centre ( suffix used ##_c##), the inner spherical shell (suffix used ##_a##) and the outer spherical shell (suffix used ##_b##).

$$V_c=\frac{q^2}{4\pi\epsilon_0}\left(\frac{-1}{a}+\frac{1}{b} \right)$$ (This ##V_c## is the potential at the centre due the other two charges). Similarly,

$$V_a=\frac{q^2}{4\pi\epsilon_0}\left(\frac{1}{a}+\frac{1}{b} \right)$$ and $$V_b=\frac{q^2}{4\pi\epsilon_0}\left(\frac{1}{b}+\frac{-1}{b} \right)$$

So, the total work required to assemble the charges is $$W=\frac{1}{2}[qV_c-qVa+qV_b]$$

This comes out to be $$W=-\frac{1}{4\pi\epsilon_0}\left( \frac{1}{a^2}\right)$$

, which is the wrong answer.

Please someone help me. I'm sure that I am missing some small detail.This problem has been driving me nuts.