Two concentric thin conducting spherical shells have radii R and 2R and carry charges +Q (inner) and -2Q (outer). The two shells are then joined by a…

Let $k=1/4\pi\varepsilon_0$. Initially $V_{\text{in}} = \dfrac{kQ}{R}-\dfrac{2kQ}{2R}=0$ and $V_{\text{out}}=\dfrac{k(Q-2Q)}{2R}=-\dfrac{kQ}{2R}$. Stored energy $U_i=\tfrac12\sum qV = \tfrac12(-2Q)\left(-\dfrac{kQ}{2R}\right)=\dfrac{kQ^2}{2R}$. After joining, all the charge $-Q$ sits on the outer shell: $U_f = \dfrac{kQ^2}{2(2R)}=\dfrac{kQ^2}{4R}$. Heat $= U_i-U_f = \dfrac{kQ^2}{4R} = \dfrac{Q^2}{16\pi\varepsilon_0 R}$. The inner shell is at zero potential initially, so its own charge does not appear in $\sum qV$ directly, though it shapes the potential of the outer shell. Joining the shells forces the inner shell's charge $+Q$ to flow outward and neutralise half of the outer charge, and the lost electrostatic energy appears as heat in the wire.