Chapter 20, Problem 1OQ

A parallel-plate capacitor is charged and then is disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled? (a) It becomes four times larger. (b) It becomes two times larger. (c) It stays the same. (d) It becomes one-half as large. (e) It becomes one-fourth as large.

To determine

Find the factor in which the energy stored in the capacitor changes when the distance between the plates is doubled.

Answer to Problem 1OQ

Option (B) It becomes two times larger.

Explanation of Solution

Write the equation for capacitance

    $C=k{\epsilon }_0\frac{A}{d}$        (I)

Here, $C$ is the capacitance, $k$ is the constant, ${\epsilon }_0$ is the permeability, $A$ is the area and $d$ is the distance between the plates.

If the distance between the plates is doubled then the above equation becomes,

    $k{\epsilon }_0\frac{A}{2d}=\frac{C}{2}$        (II)

As the charge remains constant the capacitance decreases by factor of 2.

Write the equation for potential energy stored.

    $U=\frac{Q^2}{2C}$        (III)

Here, $U$ is the potential energy stored and $Q$ is the charge.

Substitute equation II in III.

    $\begin{array}{l}U=\frac{Q^2}{2\left(C/2\right)}\\ 2U=\frac{Q^2}{2C}\end{array}$

Therefore, the potential energy becomes doubles when the distance between the plates are doubled.

Conclusion:

From the above calculation the potential energy is two times the original. Thus, option (B) is correct.