Question

A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be

-doubled

-unchanged

-quadrupled

-quartered

-halved

Answer 1

Concepts and reason

The concept required to solve this problem is energy stored in a parallel plate capacitor.

Initially, write the expression for capacitance of a parallel plate capacitor in terms of area of the plates and separation between the plates. Later, write the expression for the energy stored in a parallel plate capacitor in terms of capacitance and voltage. After that, derive expression for energy stored in the capacitor after the plate separation is doubled. Finally, compare the new energy stored with initial energy stored in the capacitor.

Fundamentals

The capacitance C of a parallel plate capacitor is given by,

$C = \frac{{{\varepsilon _0}A}}{d}$

Here, ${\varepsilon _0}$ is permittivity of frees space, $A$ is area of each plate of the capacitor, and $d$ distance between the two plates.

The expression for the energy stored in a capacitor is given by,

$E = \frac{1}{2}C{V^2}$

Here,$V$ is the voltage applied to the capacitor.

Substitute $\frac{{{\varepsilon _0}A}}{d}$ for $C$ in the equation$E = \frac{1}{2}C{V^2}$.

$E = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}$

The expression for energy stored in a capacitor is,

$E = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}$

Substitute $2d$ for $d$ in the equation$E = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}$, and derive equation for new energy$E'$stored in the capacitor after the after the plate separation is doubled.

$\begin{array}{c}\\E' = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{{\left( {2d} \right)}}} \right){V^2}\\\\ = \frac{1}{2}\left[ {\frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}} \right]\\\end{array}$

The expression for the energy stored in a parallel plate capacitor is,

$E = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}$

The expression for energy stored in the capacitor after the distance between the plates is doubled is,

$E' = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}$

Substitute $2d$ for $d$ in the equation$E' = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}$.

$\begin{array}{c}\\E' = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{{\left( {2d} \right)}}} \right){V^2}\\\\ = \frac{1}{2}\left[ {\frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}} \right]\\\end{array}$

Substitute $E$for$\frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){V^2}$.

$E' = \frac{1}{2}E$

Therefore, the option (E) is correct.

Ans:

The energy stored in the capacitor is halved if the distance between the plates is doubled.