Electric Potential: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:

By conservation of volume, $8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R = 2r$. Charge is conserved, so $Q = 8q$. The new potential $V' = \frac{kQ}{R} = \frac{k(8q)}{2r} = 4 \left(\frac{kq}{r}\right) = 4 \times 10 \text{ V} = 40 \text{ V}$.