Physics I / Electric Potential and Capacitance
Practice question · Multiple choice

A charged parallel-plate capacitor is disconnected from the battery. You pull the plates farther apart. What happens to the voltage between them?

Hints
  1. Disconnected plates trap their charge. For large plates, the field depends only on the charge density, not on the gap.
  2. V=EdV = Ed: constant EE, growing dd. You also do work against the plates’ attraction, where does that work go?
Show the answer

A. It increases, same field over a larger gap

Why

With QQ trapped, E=σ/ε0E = \sigma/\varepsilon_0 stays fixed, so V=EdV = Ed grows with the gap. The work of pulling the plates apart is stored as extra field energy, you can literally raise a voltage with your bare hands.

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