I've been reviewing electrostatics using an old exam and I stumbled upon this question:

Calculate the amount of work required to assemble a net charge of $+Q$ on a spherical conductor of radius $R$. If an additional charge of $-Q$ were to be assembled on a concentric spherical conductor of radius $R+a$,what amount of work would the entire process require?

Now the first part is not that difficult, we just do:

$$\vec E = \frac{Q}{4 \pi \epsilon_0 R^2} \hat r \, \text{(From Gauss's Law)}$$

$$\begin{align} W & = \frac{\epsilon_0}{2} \int E^2 \, d\tau \\ & = \left(\frac{\epsilon_0}{2}\right) \left(\frac{Q^2}{(4 \pi \epsilon_0)^2}\right) \int d \Omega \int_R^{\infty} \frac{1}{R'^{4}} R'^{2} dR'\\ & = \frac{4\pi Q^2}{32 \pi^2 \epsilon_0} \frac{1}{R} \\ &= \frac{Q^2}{8 \pi \epsilon_0 R} \\ \end{align}$$

But for the second part, according to a solution that a friend of mine gave me, only thing that we need to do to calculate the total work is to do:

$$\begin{align} W_{tot} & = \frac{\epsilon_0}{2} \int E^2 d \tau \\ & = \frac{4 \pi Q^2}{32 \pi^2 \epsilon_0} \int_R^{R+a} \frac{1}{R'^2} dR' \\ & = \frac{Q^2}{8 \pi \epsilon_0} \left(\frac{1}{R} - \frac{1}{R+a}\right) \\ \end{align}$$

But according to equation $(2.47)$ of Griffiths, total work should be equal to:

$$\begin{align} W_{tot} & = \frac{\epsilon_0}{2} \int (E_1+E_2)^2 d\tau \\ & = \frac{epsilon_0}{2} \int (E_1^2 + E_2^2 + 2E_1 \cdot E_2) d\tau \\ & = W_1 + W_2 + \epsilon_0 \int E_1 \cdot E_2 d \tau \\ \end{align}$$

Wherein for this case $W_1$ is the work required for a sphere of radius $R$ as shown earlier, and $W_2$ is the work required for a sphere of radius $R+a$. Is the first method correct?

  • $\begingroup$ The total work $W_{tot}$ caculated by using the first method has a wrong sign, it should be $-(\frac{1}{R+a}-\frac{1}{R})$, i.e.$\frac{1}{R}-\frac{1}{R+a}$. $\endgroup$
    – Wang Yun
    Jan 4, 2016 at 6:26
  • $\begingroup$ @StephenWong Oh sorry must've missed that one, but is the method correct tho? $\endgroup$
    – mopy
    Jan 4, 2016 at 6:46
  • $\begingroup$ I think my answer would help you. $\endgroup$
    – Wang Yun
    Jan 4, 2016 at 7:08

1 Answer 1


Answer:The two methods are both correct.

As I have suggested in the comment area, total work calculated by using the first method should be $$W_{tot}=\frac{Q^2}{8\pi\epsilon_0}(\frac{1}{R}-\frac{1}{R+a}), $$ since $\int \frac{1}{r^2}dr=-\frac{1}{r}+\text{Constant}$.

Next we will calculate the total work by the second method, i.e. the equation $(2.47)$ of Griffiths. As indicated in the problem, we have \begin{align} \mathbf{E}_1 & =\frac 1 {4\pi\epsilon_0}\frac{Q}{r^2}\hat{\mathbf{r}},\ \text{while}\ r\ge R\\ \mathbf{E}_2 & =-\frac 1 {4\pi\epsilon_0}\frac{Q}{r^2}\hat{\mathbf{r}},\ \text{while}\ r\ge R+a \end{align}

So, \begin{align} E_1^2 & =\frac{Q^2}{16\pi^2\epsilon_0^2r^4}\\ E_2^2 & =\frac{Q^2}{16\pi^2\epsilon_0^2r^4}\\ E_1\cdot E_2 & =-\frac{Q^2}{16\pi^2\epsilon_0^2r^4} \end{align}

And \begin{align} W_1 & =\frac{\epsilon_0}{2}\int E_1^2d\tau\\ & =\frac{Q^2}{8\pi\epsilon_0}\int_R^\infty\frac{1}{r^2}dr\\ & =\frac{Q^2}{8\pi\epsilon_0R} \end{align} By using same method, we get $W_2$ as foolow, $$ W_2=\frac{Q^2}{8\pi\epsilon_0(R+a)} $$

Finally, the cross term is that \begin{align} \epsilon_0 \int \mathbf{E}_1 \cdot \mathbf{E}_2 d\tau & =-\frac{Q^2}{4\pi\epsilon_0}\int_{R+a}^\infty \frac{1}{r^2}dr\\ & =-\frac{Q^2}{8\pi\epsilon_0}\frac{2}{R+a} \end{align}

Then we add them all up, we have, \begin{align} W_{tot} & =W_1+W_2+\epsilon_0\int \mathbf{E}_1\cdot\mathbf{E}_2d\tau \\ & =\frac{Q^2}{8\pi\epsilon_0}\frac{1}{R}+\frac{Q^2}{8\pi\epsilon_0}\frac{1}{R+a}-\frac{Q^2}{8\pi\epsilon_0}\frac{2}{R+a}\\ & =\frac{Q^2}{8\pi\epsilon_0}(\frac{1}{R}-\frac{1}{R+a}) \end{align}

Conclusion: The results are the same by two methods.In the first method, when we wrote the formula of $W_{tot}$, the electric field $E$ is the final field after used superposition princeple. In the second, we also used superposition princeple, but we wrote it in the form of $W$ explicitly. I mean that the two methods are the same, but have different forms.

  • $\begingroup$ Obviously the second method will work since it's the recommended one from Griffiths, but why did the first one work? Do you have any insights and situations wherein it will work and will not work? $\endgroup$
    – mopy
    Jan 4, 2016 at 9:59
  • $\begingroup$ The first method is the most basic, and the second is derived from the first according superposition princeple of electric field. $\endgroup$
    – Wang Yun
    Jan 4, 2016 at 10:21
  • $\begingroup$ @Aldon I modified my answer again, I hope that would help you. $\endgroup$
    – Wang Yun
    Jan 5, 2016 at 1:10
  • $\begingroup$ @StephenWong Do you know why it is that in Griffiths Intro to Electrodynamics Fourth edition, he gets your answer (but negative) for problem 2.60? $\endgroup$
    – user100411
    Aug 1, 2016 at 21:02

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