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?