1. **Problem statement:** A solid conducting sphere with charge $Q$ is surrounded by an uncharged conducting hollow spherical shell. The potential difference between the solid sphere's surface and the outer surface of the shell is $V$. If the shell is given a charge of $-4Q$, find the new potential difference between the same two surfaces. 2. **Key concepts:** - The potential on the surface of a charged conducting sphere of radius $R$ with charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$. - For a conducting shell, the charge resides on the outer surface, and the potential inside the shell is constant. - The potential difference between two points is the difference in their potentials. 3. **Initial setup:** - Let the radius of the solid sphere be $R_1$ and the outer radius of the hollow shell be $R_2$. - Initially, the shell is uncharged, so the potential on the outer surface of the shell is due only to the solid sphere's charge $Q$. - Potential on solid sphere surface: $V_1 = \frac{1}{4\pi\epsilon_0} \frac{Q}{R_1}$. - Potential on outer shell surface: $V_2 = \frac{1}{4\pi\epsilon_0} \frac{Q}{R_2}$. - Initial potential difference: $V = V_1 - V_2 = \frac{1}{4\pi\epsilon_0} Q \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. 4. **After charging the shell with $-4Q$:** - The shell now has charge $-4Q$ on its outer surface. - The potential on the solid sphere surface is due to its own charge $Q$ plus the shell's charge $-4Q$. - Potential on solid sphere surface: $$V_1' = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{R_1} + \frac{-4Q}{R_2} \right) = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{R_1} - \frac{4Q}{R_2} \right)$$ - Potential on outer shell surface: $$V_2' = \frac{1}{4\pi\epsilon_0} \frac{-4Q}{R_2}$$ 5. **New potential difference:** $$\Delta V' = V_1' - V_2' = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{R_1} - \frac{4Q}{R_2} \right) - \frac{1}{4\pi\epsilon_0} \left( \frac{-4Q}{R_2} \right) = \frac{1}{4\pi\epsilon_0} \frac{Q}{R_1}$$ 6. **Interpretation:** The new potential difference is simply the potential on the solid sphere surface due to its own charge $Q$ because the shell's charge contributions cancel out. **Final answer:** $$\boxed{\Delta V' = \frac{1}{4\pi\epsilon_0} \frac{Q}{R_1}}$$