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).

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

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.

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

To solve these problems, we need to use the concepts of capacitance and the relationship between potential difference, charge, and potential energy in a capacitor. Let's go step by step:

Situation I: Distance between the plates is quadrupled (d -> 4d)
a. The potential difference (V) across the plates of a capacitor is directly proportional to the distance (d) between the plates. Therefore, if the distance is quadrupled, the potential difference will also be quadrupled. Thus, the potential difference measured across the plates (VC) now would be four times the original potential difference.

b. The potential energy (U) stored in a capacitor is given by the formula: U = (1/2) * Q^2 / C, where Q is the charge stored in the capacitor and C is the capacitance. Since the distance between the plates has changed, the capacitance will also change. However, in this case, the charge is not changed. So, the potential energy stored in the capacitor (UC) will remain the same as the original potential energy.

Situation II: Distance between the plates is divided by 4 (d -> d/4)
c. In this situation, the capacitor is still connected to the battery while the distance between the plates is changed. According to the formula V = Q / C, where V is the potential difference, Q is the charge, and C is the capacitance, the potential difference is inversely proportional to the distance between the plates. Therefore, if the distance is divided by 4, the potential difference will also be divided by 4. Hence, the potential difference measured across the plates (VC) now would be one-fourth of the original potential difference.

d. The potential energy stored in a capacitor is still given by the formula U = (1/2) * Q^2 / C, where Q is the charge stored and C is the capacitance. Again, in this case, the charge is not changed, but the capacitance will be affected by the change in distance. Since the distance is divided by 4, the capacitance will quadruple, resulting in a potential energy (UC) that is four times the original potential energy.

Remember that these answers assume that only the distance between the plates is changed, and the shape, area, and charge of the plates remain the same.