A spherical conducting shell of radius $R_1$ with a uniform surface charge $q$ has a point charge $q

Question

A spherical conducting shell of radius $R_1$ with a uniform surface charge $q$ has a point charge $q_0$ located at its geometric centre. The shell is slowly expanded to a new radius $R_2$ ($R_2 > R_1$). Which of the following statements is/are CORRECT?

Professor PC Thomas Classes · Answer

The work done by electric forces is $W_e = -\Delta U = U_i - U_f$. The total energy of the system $U$ is the sum of the self-energy of the shell and the interaction energy between the shell and the point charge. 
The self-energy of the shell is $U_{self} = \frac{q^2}{8\pi\epsilon_0 R}$. The change is $\Delta U_{self} = \frac{q^2}{8\pi\epsilon_0} (\frac{1}{R_2} - \frac{1}{R_1})$. Since $R_2 > R_1$, this change is negative, meaning self-energy decreases by $\frac{q^2}{8\pi\epsilon_0}(\frac{1}{R_1} - \frac{1}{R_2})$. (B is incorrect)
The potential at the shell due to $q_0$ is $\frac{q_0}{4\pi\epsilon_0 R}$, so the interaction energy is $U_{int} = q V = \frac{q q_0}{4\pi\epsilon_0 R}$. 
The change is $\Delta U_{int} = \frac{q q_0}{4\pi\epsilon_0} (\frac{1}{R_2} - \frac{1}{R_1})$, meaning it decreases by $\frac{q q_0}{4\pi\epsilon_0}(\frac{1}{R_1} - \frac{1}{R_2})$ (C is correct). 
The total work done is $W_e = -\Delta U_{self} - \Delta U_{int} = \frac{q}{4\pi\epsilon_0}\left(\frac{q}{2} + q_0\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ (D is correct). 
Because $1/R_1 > 1/R_2$, if both $q$ and $q_0$ are positive, this total work is undeniably positive (A is correct).