For the capacitor described above with the same plate area, what is the potential difference across the plates if the plate separation is 3.00 mm?

• A. 300 V
• B. 330 V
• C. 450 V
• D. 600 V

Answer: (B)

The key idea is how a parallel-plate capacitor behaves when the plate separation changes while the plate area stays the same. The capacitance is C = ε0 A / d, so the potential difference for a given charge is V = Q / C = Q d / (ε0 A). This means, with the same area and the same charge on the plates, the voltage grows linearly with the separation d.

If the separation is set to 3.00 mm, you plug that into V = Q d /(ε0 A). The numbers for Q and A come from the given problem setup, and doing that substitution yields 330 V. In short, increasing the distance while keeping the charge fixed lowers the capacitance and raises the voltage by the same factor as the distance increases, which is why 330 V is the correct result for the 3.00 mm gap.