A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. The plates are then pulled apart. Explain whether each of the following quantities increases, decreases, or remains the same as the distance between the plates increases for each of these cases
.
(a) the capacitance of the capacitor(b) the potential difference between the plates(c) the electric field between the plates(d) the electric potential energy stored by the capacitor

a. Decreases

b. Increases

c. Remains the same

d. Increases

Explanation:

a. Capacitance is given by c= Ak€/d

where A is conductivity plate with Area

K is a constant

€ is dielectric with permittivity.

d is the distance

b. Potential difference is given by

V = Ed, since, the electric field remains the

same, the potential diterence also increases with increase in distance.

Since the capacitance depends upon the distance, and all the other factors are kept constant, the capacitance decreases.

c. Electric field remains the same because charge on the

plate remains the same.

d. since electric field remains the same and capacitance decreases, the energy increases.

E= 1/2c * Q^2

i think the answer is a, hope this : )

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