Calculating the amount of work done to assemble a net charge on a sphere

Question

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?

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.