If a parallel plate capacitor is fully charged by a battery?

Suppose a parallel plate capacitor (with capacitance C1) is fully charged (to a value Qo) by a battery. The battery (which supplies a potential difference of Vo) is then disconnected. If the plates of the capacitor are then moved ƒᴀʀтher apart (the separation distance d between the plates is doubled), describe quantitatively what happens to:

Answer 1

1) The capacitance is halved: C = ε0*A/d.

2) Potential difference doubles V = Q/C Q doesn’t change, so if C is halved, V is doubled.

3) Charge cannot go anywhere so it must stay the same

4) Energy = 0.5*C*V² so if C is halved and V doubled, energy is doubled. You might ask where does this additional energy come from? There is a force of attraction between the plates, and moving them apart takes energy.

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