<s>[INST] Calculating the amount of work done to assemble a net charge on a sphere 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?
 [/INST] 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.
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