A capacitor is charged to 40.0 C and the potential difference across the plates is 15.0 V. What is the potential difference across its plates when the energy stored in the capacitor is 50.0 mJ?

What you are given and what changes

You are told one charging condition ($Q=40.0\,\text{C}$ at $V=15.0\,\text{V}$), which lets you determine the capacitor’s fixed capacitance $C$. Then you use the energy formula to find the new voltage when the stored energy is $50.0\,\text{mJ}$.

Find the capacitance from $Q=CV$

Use

$$C = \frac{Q}{V}$$

Substitute the given values:

$$C = \frac{40.0\,\text{C}}{15.0\,\text{V}} = 2.666\ldots\,\text{F} \approx 2.67\,\text{F}$$

Use stored energy to solve for the voltage

The energy stored in a capacitor is

$$U = \frac{1}{2}CV^2$$

Convert energy: $50.0\,\text{mJ} = 0.0500\,\text{J}$.

Solve for $V$:

$$V = \sqrt{\frac{2U}{C}}$$

Plug in:

$$V = \sqrt{\frac{2(0.0500)}{2.67}} = \sqrt{0.0375} \approx 0.19365\,\text{V} \approx 0.194\,\text{V}$$

Quick reasonableness check

Since $U \propto V^2$, a small energy like $0.0500\,\text{J}$ implies a much smaller voltage than $15\,\text{V}$ for such a large capacitance ($\sim 2.7\,\text{F}$), so a value under $1\,\text{V}$ makes sense.