If you double the plate area of a capacitor while keeping the plate separation and dielectric constant constant, what happens to the capacitance?

If you double the plate area of a capacitor while keeping the plate separation and dielectric constant constant, what happens to the capacitance?

• A. It doubles
• B. It increases by a factor of four
• C. It remains the same
• D. It becomes zero

Answer: (A)

Capacitance scales with how much plate area is facing each other. For a parallel-plate capacitor, C = ε A / d. If the dielectric (ε) and the plate separation (d) stay the same and you double the area (A), the capacitance doubles: C_new = ε (2A) / d = 2C_old. Physically, more surface area means more charge can be stored at the same voltage for the field between the plates. So the correct outcome is that the capacitance increases by a factor of two.

Capacitance scales with how much plate area is facing each other. For a parallel-plate capacitor, C = ε A / d. If the dielectric (ε) and the plate separation (d) stay the same and you double the area (A), the capacitance doubles: C_new = ε (2A) / d = 2C_old. Physically, more surface area means more charge can be stored at the same voltage for the field between the plates. So the correct outcome is that the capacitance increases by a factor of two.