Question

A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while the capacitor remains connected to the battery?
(a) It remains the same
(b) It is doubles
(c) It decreases by a factor of 2
(d) It decreases by a factor of 4
(e) It increases by a factor of

Answer 1

Answer:

(c) It decreases by a factor of 2

Explanation:

Since the capacitor is still connected to the power source, the potential difference remain the same even when the distance is a doubled.

The energy stored in a capacitor can be written as:

E = (1/2)CV^2 .....1

And the capacitance of a capacitor is inversely proportional to the distance between the two plates of the capacitor.

C = kA/d ....2

Therefore, when d doubles, and every other determinant of capacitance remains the same, the capacitance is halved.

Cf = kA/2d = C/2

Cf = C/2

Since the capacitance has been halved and potential difference remains the same, the energy stored would also be halved since the energy stored in the capacitor is directly proportional to the capacitance.

Ef = (1/2)(Cf)V^2

Ef = (1/2)(C/2)V^2 = [(1/2)CV^2]/2

Ef = E/2

Where;

E and Ef are the initial and final energy stored in the capacitor respectively

C and Cf are the initial and final capacitance of the capacitor.

d is the distance between the plates

A is the area of plates

k is the permittivity of dielectrics

Therefore the energy stored in the capacitor is decreased by a factor of 2, when the distance is doubled.

Answer 2

Answer:

(c)  As 'd' becomes doubled, energy decreases by the factor of 2

Explanation:

Energy stored in a parallel plate capacitor is given by:

$U=\frac{1}{2}CV^2\\\\C=\frac{A\epsilon_{o}}{d}\\\\then\\\\U=\frac{1}{2}\frac{A\epsilon_{o}}{d}V^2--(1)$

As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.