Question

A parallel plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the plates are pulled away from each other, increasing their separation, what happens to the amount of charge on the plates?

a. The amount of the charge decreases, because the capacitance increases.
b. Nothing happens; the amount of charge stays the same.
c. The amount of the charge increases, because the capacitance increases.
d. The amount of the charge increases, because the capacitance decreases.
e. The amount of the charge decreases, because the capacitance decreases.

Answer 1

When the plates of a parallel plate capacitor are pulled away from each other, the amount of charge stays the same.

Step-by-step explanation:

When the plates of a parallel plate capacitor are pulled away from each other, increasing their separation, the amount of charge stays the same (option b). This is because the charge on the plates is determined by the voltage and capacitance, and changing the distance between the plates does not affect these factors.

The capacitance of a parallel plate capacitor is given by the equation C = εA/d, where ε is the permittivity of the material between the plates, A is the area of each plate, and d is the separation between the plates. As the separation increases, the capacitance actually decreases, but this does not affect the amount of charge on the plates.

Answer 2

e. The amount of the charge decreases, because the capacitance decreases.

Step-by-step explanation:

The capacitance of a parallel plate capacitor is given by the formula:

`C=(\epsilon A)/(d)`

where
`\epsilon` is the permittivity of the medium between the plates, A the area of the plates and d the separation between them, so if we increase the separation d the capacitance C will decrease.

The relationship between the charge on the plate Q and the voltage applied V (the potential difference between the plates) is given by:

`Q=CV`

This means that if while keeping the potential difference between the plates V constant the capacitance C decreases then the charge Q will decrease as well.