Question

After a parallel-plate capacitor has been fully charged by a battery, the battery is disconnected and the plate separation is increased. Which of the following statements is correct? Please explain in detail why the staement is correct!
A) The energy stored in the capacitor increases.
B) The charge on the plates increases.
C) The charge on the plates decreases.
D) The potential difference between the plated decreases.
E) The energy stored in the capacitor decreases.

Answer 1

Answer:

C

Explanation:

Using the two formulas q =CV and C =∈A/d where q = charge, C = Capacitance, V =Voltage, ∈=permittivity of dielectric, A = area of plates and d = distance between plates. Increasing the plate separation decreases the Capacitance thereby decreasing the charge also.

Answer 2

Answer:

A) The energy stored in the capacitor increases.

Explanation:

For a capacitor fully charged by battery, when disconnected from battery and the plate separation is increased. The charge on the plate remain constant because there is no where for it to go( it has been disconnected from battery). But the capacitance would decrease, while also the potential difference would increase.

Q = CV ....1

Q is the charge, C is capacitance, V is the potential difference.

The energy stored in a capacitor is given by:

E = 1/2 CV^2

E = 1/2 QV .......2

E is the energy stored in the capacitor,

Therefore since Q remain constant and V increases when the distance between the plates is increased, then according to the equation 2 above the energy stored in the capacitor increases.