If the distance between the plates is halved, what happens to the capacitance?

If the distance between the plates is halved, what happens to the capacitance?

• A. It doubles
• B. It halves
• C. It remains unchanged
• D. It increases by a factor of four

Answer: (A)

Capacitance of a parallel-plate capacitor is inversely proportional to the distance between the plates. For a capacitor with area A and dielectric constant ε, C = εA / d. Halving the separation makes the denominator d/2, so the capacitance becomes C' = εA / (d/2) = 2εA / d = 2C. Therefore, the capacitance doubles. This isn’t a fourfold change because the dependence is 1/d, not 1/d^2. The area and dielectric are unchanged, so the only change is the separation.

Capacitance of a parallel-plate capacitor is inversely proportional to the distance between the plates. For a capacitor with area A and dielectric constant ε, C = εA / d. Halving the separation makes the denominator d/2, so the capacitance becomes C' = εA / (d/2) = 2εA / d = 2C. Therefore, the capacitance doubles. This isn’t a fourfold change because the dependence is 1/d, not 1/d^2. The area and dielectric are unchanged, so the only change is the separation.