A parallel capicitor is made of two plates with cross sectional area A which are separated by a distance d (with a vacuum between the plates). The capacitor is connected to a battery until a charge Q is stored on it and a total potential difference VC is measured across the plates (the potential energy stored at this time is UC).

Question

Situation I: The capacitor is disconnected from the battery, and its plates moved so the distance between them is quadrupled.
a. What is the potential difference measured across the plates now (as a multiple of the original potential difference)?
VC
b. What is the potential energy stored in the capacitor now (as a multiple of the original potential energy stored)?
UC
Situation II: Instead of the above, the capacitor is left connected to the battery, and its plates moved so the distance between them is divided by 4.
c. What is the potential difference measured across the plates now (as a multiple of the original potential difference)?
VC
d. What is the potential energy stored in the capacitor now (as a multiple of the original potential energy stored)?
UC

Answer 1

I. The charge is constant. What happens to C when d>>4d?
It took work to move the plates apart, so PE must be less. PE=1/2 QV^2

II. V is constant. What happened to C? Now, figure q stored.

Answer 2

C=Q/V
C=εε₀A/d
Situation I: The capacitor is disconnected from the battery, => Q=const
(a)
V=Q/C=Qd/εε₀A
d₁=4d=>
V₁=Q/C₁=Qd₁/εε₀A=4 Qd/εε₀A=4V
(b)
U=CU²/2=Q²d/2εε₀A
U₁=C₁U₁²/2=Q²d₁/2εε₀A=
=Q²4d₁/2εε₀A=4U.

Situation II. The capacitor is left connected to the battery => V=const

(a) V₂=V
(b) U₂=C₂V²/2= εε₀AV²/d₂=
=4 εε₀AV²/d=4U