# Work done in moving a charge to infinity

1. May 23, 2016

### Aniruddha@94

1. The problem statement, all variables and given/known data
(Not for homework/assignment. Just doing problems for practice)
This is from Griffiths Introduction to Electrodynamics, 4th 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: $(q^2/4\pi\epsilon_0)(1/a)$ ]
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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
2. Relevant equations
$$W=\frac{\epsilon_0}{2}∫E^2d\tau$$
$$V(r_1)-V(r_2)=\frac W q$$
$$W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i)$$

3. 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
$$\frac{q^2}{8\pi\epsilon_0}\left[ \int_0^a\frac{1}{r^2}dr+\int_b^∞\frac{1}{r^2}dr\right]$$
$$=\frac{q^2}{8\pi\epsilon_0}\left[\frac{-1}{a}+\frac{1}{0}-\frac{1}{∞}+\frac{1}{b}\right]$$
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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.

2. May 23, 2016

### TSny

The 4 in the denominator should be 8.

The infinity is due to the infinite “self energy” of a point charge. When you remove the point charge to infinity, that infinite self energy will still exist. But you are only interested in the difference ΔW between initial and final states.

Another way to think about it, using this approach, is to consider the difference between the electric field distribution of the point charge with the spherical shell in place and the electric field of the point charge isolated by itself. You can see that ΔW is due to just the absence of E in the range a < r < b for the initial configuration. So, you can determine ΔW by just an integration of E2 of the point charge over this region. It should give the same answer as your integration.
In the equation $W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i)$ , $V(\textbf{r}_i)$ is the total potential at the location of $q_i$, including the contribution from $q_i$ itself. So inside the parentheses in your expression for $V_c$ there should also be an infinite terms, 1/0, representing the potential of the point charge at the location of the point charge. But this is related to the infinite self energy and can be ignored. So, your expression for $V_c$ is OK, except q shouldn't be squared, right?

[EDIT: To avoid worrying about infinite self energy of point charges, the equation $W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i)$ is used where $V(\textbf{r}_i)$ is the potential at the location of $q_i$ due to all charges other than $q_i$. So, your expression for $V_c$ then follows just as you derived it (except for the square on q). For continuous charge distributions, $q_i$ in the formula for $W$ is replaced by an infinitesimal element of charge $dq$ and the sum is replace by an integral.]
This is missing the potential at r = a due to the charge –q at r = a. (This is not infinite.) Again, should q be squared?
This needs to be fixed similarly to fixing $V_a$.

Last edited: May 23, 2016
3. May 23, 2016

### Aniruddha@94

Sorry for these, typing errors. I didn't mean to write $q^2$.
I now understand the thing about infinite self energy. It's like we're trying to find the energy required to put together a point charge and it comes out to be infinity, right? And I got the correct answer using your suggestion (i.e, energy stored in the field b/w the spheres of radius $a$ and $b$). But, in my first method, can I simply leave out the $\frac{1}{0}$ part? (Although I feel doing that isn't mathematically correct, and it gives me an unwanted negative sign ).
Also, I've got the error in the second method. For the relation $W = \frac{1}{2} \sum_{i=1}^n q_i V(\textbf{r}_i)$ , when we're dealing with point charges we leave out the charge at the point we consider, but when it comes to continuous distributions we take the potential of that body too, right?
So, now we get $$V_c=\frac{q}{4\pi\epsilon_0}\left(\frac{-1}{a}+\frac{1}{b} \right)$$ $$V_a=\frac{q}{4\pi\epsilon_0}\left(\frac{1}{a}-\frac{1}{a}+\frac{1}{b} \right)$$ and
$$V_b=\frac{q}{4\pi\epsilon_0}\left(\frac{1}{b}+\frac{-1}{b}+\frac{1}{b} \right)$$
So, $W=\frac{1}{2}[qV_c-qVa+qV_b]$ gives $$W=\frac{q^2}{4\pi\epsilon_0}\left( \frac{-1}{a}+\frac{1}{b}\right)$$
Still this negative sign is unaccounted for.

4. May 23, 2016

### TSny

Yes, you can just leave it out. The work required to move the charge from the center of the shell to infinity is the difference $W_f - W_i$, where $W_i$ is the electrostatic energy initially and $W_f$ is the electrostatic energy finally. Both $W_i$ and $W_f$ will contain the infinite self energy 1/0. But it cancels out in taking the difference. You could model the point charge by a very small sphere of radius $\epsilon$. You would then get a term containing$1/\epsilon$ in both $W_i$ and $W_f$ which cancels in taking the difference. The overall negative sign that you are getting will be taken care of when you calculate $W_f - W_i$.
Yes, that's right.
Your expression represents the initial energy $W_i$. It looks good,except the 4 should be an 8 in your last equation. You still need to think about $W_f$ and then find work = ΔW = Wf - Wi.

Last edited: May 23, 2016
5. May 23, 2016

### ehild

The problem can be solved by calculating the whole energy of the electric field using energy density (ε/2)E2 . Initially, the q charge was inside the shell, and the electric field was E=kq/r2, except inside the wall of the shell, a<r<b. When the charge is removed to infinity, the electric field is kq/r2 everywhere around the charge.The difference of energies is equal to the work done.

6. May 23, 2016

### TSny

Hi, ehild! Yes, this approach was mentioned in post #2. And I believe Aniruddha said in post #3 that he was able to get the answer this way (except for the overall sign?).

Last edited: May 23, 2016
7. May 23, 2016

### ehild

Yes, I see now.

8. May 23, 2016

### Aniruddha@94

Aaarrgh! Stupid typo again (sorry about those, I'm using these codes for the first time). Anyways, I don't understand what $W_f$ should be here. I got what it is in the first method.
Yes, I've got the answer. The overall sign depends upon whether I take $\int_a^b$ or $\int_b^a$ . Should there any preference?

9. May 23, 2016

### TSny

If a spring is initially stretch so that it has a potential energy Ui and then you stretch it some more until the potential energy is Uf, the work you do is ΔU = Uf - Ui. Likewise, when the point charge is at the center of the shell, the system has an electrostatic potential energy Wi. After moving the point charge to infinity, the system has a new potential energy Wf. (You can use the symbol U instead of W here if you wish.) The work required to move the charge is ΔW = Wf -Wi. If the self energy of the point charge is ignored, then Wf = 0. Then the work required is just ΔW = 0 - Wi = -Wi. That is, the work is the negative of the initial potential energy.
Yes, you need to take $\int_a^b$. The integral comes from using $4 \pi r^2 dr$ as the volume element. A volume element should be a positive number. Thus, $dr$ needs to be positive. Integrating from $a$ to $b$ makes $dr$ positive.

10. May 24, 2016

### Aniruddha@94

Oh! I see it now. Thanks a lot for explaining it.