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, $27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R = 3r$. Total charge $Q = 27q$. The new potential is $V_{new} = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9V_{old}$. Thus, $V_{new} = 9 \times 220\text{ V} = 1980\text{ V}$.