If the plate separation d of a parallel-plate capacitor in vacuum is doubled, what happens to the capacitance?

If the plate separation d of a parallel-plate capacitor in vacuum is doubled, what happens to the capacitance?

• A. Capacitance doubles
• B. Capacitance halves
• C. Capacitance remains the same
• D. Capacitance becomes zero

Answer: (B)

Capacitance depends on how strongly the plates can store charge for a given voltage. For a parallel-plate capacitor in vacuum, C = ε0 A / d. If the separation d is doubled, the denominator doubles and the whole fraction becomes half as large. So the capacitance becomes half of its original value. Intuitively, increasing the distance between the plates weakens the electric coupling between them, so for the same charge, you get a lower voltage, or equivalently, for the same voltage you store less charge. Edge effects are neglected in this simple formula, which is why the inverse relationship with distance holds.

Capacitance depends on how strongly the plates can store charge for a given voltage. For a parallel-plate capacitor in vacuum, C = ε0 A / d. If the separation d is doubled, the denominator doubles and the whole fraction becomes half as large. So the capacitance becomes half of its original value.

Intuitively, increasing the distance between the plates weakens the electric coupling between them, so for the same charge, you get a lower voltage, or equivalently, for the same voltage you store less charge. Edge effects are neglected in this simple formula, which is why the inverse relationship with distance holds.