Question

A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If ‘$x$’ is the separation between the plates, then the time rate of change of electrostatic energy of the capacitor is proportional to:
A. ${x^2}$
B. $x$
C. ${x^{ - 1}}$
D. ${x^{ - 2}}$

Answer 1

Hint:Here, we will use the formula of the energy of the capacitor to calculate the time rate of change of electrostatic energy. For this, we will take the first derivative of the energy of the capacitor. Also, we will use the formula of capacitance to use it in the formula of energy.

Complete step by step answer:
A parallel plate capacitor is a setup in which two parallel plates are connected across a battery. These plates are charged and an electric field is induced between the plates. A typical parallel plate capacitor consists of two plates having an area $$A$$ and these plates are separated by a distance $d$ . Capacitance is the obstruction of the parallel plate capacitor to store energy and is given by
$C = k{\varepsilon _0}\dfrac{A}{d}$
As there is no dielectric material in the capacitor, therefore, there will be no relative permittivity. Also, $x$ is the separation between the plates, therefore, the capacitance will become
$C = {\varepsilon _0}\dfrac{A}{x}$
Now, the energy of the capacitor is given by
$U = \dfrac{1}{2}C{V^2}$
Now, putting the value of $C$ in the above equation. Also, as the battery is attached to the parallel plates capacitor, the potential $V$ in the capacitor will be constant.
Therefore, the energy $U$ is given by
$U = \dfrac{1}{2} \times {\varepsilon _0}\dfrac{A}{x} \times {V^2}$
$ \Rightarrow \,U = \dfrac{1}{2}{\varepsilon _0}A{V^2} \times \dfrac{1}{x}$
Now, taking ${\varepsilon _0}$ , $A$ and $V$ constant, we get
$U \propto \dfrac{1}{x}$
Now, taking the first derivative of energy $U$ , we get
$\dfrac{{dU}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{1}{2}{\varepsilon _0}{V^2}\dfrac{A}{x}} \right)$
$ \Rightarrow \,\dfrac{{dU}}{{dt}} = \dfrac{1}{2}{\varepsilon _0}A{V^2}\dfrac{d}{{dt}}\left( {\dfrac{1}{x}} \right)$
$ \Rightarrow \,\dfrac{{dU}}{{dt}} = \dfrac{1}{2}{\varepsilon _0}A{V^2}\left( {\dfrac{1}{{ - {x^2}}}\dfrac{{dx}}{{dt}}} \right)$
Again, taking ${\varepsilon _0}$ , $A$ and $V$ constant, we get
$\therefore\dfrac{{dU}}{{dt}} \propto \dfrac{1}{{ - {x^2}}}$
Therefore, the time rate of change of the electrostatic energy of the capacitor is proportional to $\dfrac{1}{{ - {x^2}}}$.

Hence, option D is the correct option.

Note:The energy stored in a capacitor is the electrostatic potential energy which is related to charge $Q$ and voltage $V$ between the capacitor plates. A charged capacitor always stores energy in the form of an electric field between the plates. When we disconnect the capacitor from the battery, the energy will remain in the field in the space between its plates.