Question

Question: (II) A parallel-plate capacitor has fixed charges \({\bf{ + Q}}\) and \( - {\bf{Q}}\). The separation of the plates is then halved. (a) By what factor does the energy stored in the electric field change? (b) How much work must be done to reduce the plate separation from d to \(\frac{{\bf{1}}}{{\bf{2}}}{\bf{d}}\)? The area of each plate is A.

Short answer

(a) The energy stored in the electric field is reduced by a factor\(\frac{1}{2}\).

(b) The work must be done to reduce the separation of the plates from \(d\) to \(\frac{1}{2}d\) is \( - \frac{{{Q^2}d}}{{4{\varepsilon _0}A}}\).

Step-by-step solution

Understanding the energy stored in a capacitor

A capacitor is a charge and energy storage device. The stored energy is proportional to the square of the magnitude of the charge on the plates.

The expression for stored energy in the capacitor is given as:

\({\rm{PE}} = \frac{{{Q^2}}}{{2C}}\) … (i)

Here, C is the capacitance and Q is the charge on the capacitor plates.

Given data

The separation between the plates reduces from d to \(\frac{d}{2}\).

(a) Determination of the change in stored energy

The initial capacitance of the capacitor is given as:

\(C = \frac{{{\varepsilon _0}A}}{d}\)

Here,\({\varepsilon _0}\)is the permittivity of free space, A is the plates’ area and d is the separation between the plates.

When the separation reduces, the capacitance of the capacitor is,

\(\begin{aligned}{c}C' &= \frac{{{\varepsilon _0}A}}{{\frac{1}{2}d}}\\ &= 2\frac{{{\varepsilon _0}A}}{d}\\ &= 2C\end{aligned}\)

From equation (i), the energy stored in a capacitor is inversely proportional to the capacitance when the charge is fixed.

Since the capacitance becomes double the initial value, the energy stored in the capacitor becomes half of the initial value.

Thus, the energy stored in the electric field is reduced by a factor \(\frac{1}{2}\).

(b) Determination of the work done to make this separation 

The initial potential energy stored in the capacitor is,

\({\rm{PE}} = \frac{{{Q^2}}}{{2C}}\)

The final potential energy stored in the capacitor is,

\(\begin{aligned}{c}{\rm{PF'}} &= \frac{{{Q^2}}}{{2C'}}\\ &= \frac{{{Q^2}}}{{2 \times 2C}}\\ &= \frac{{{Q^2}}}{{4C}}\end{aligned}\)

The work done to make this change equals the change in potential energy.

Then the work done is,

\(\begin{aligned}{c}W &= {\rm{PE'}} - {\rm{PE}}\\ &= \frac{{{Q^2}}}{{4C}} - \frac{{{Q^2}}}{{2C}}\\ &= - \frac{{{Q^2}}}{{4C}}\end{aligned}\)

Substitute the value of capacitance in the above expression.

\(\begin{aligned}{c}W &= - \frac{{{Q^2}}}{{4 \times \left( {\frac{{{\varepsilon _0}A}}{d}} \right)}}\\ &= - \frac{{{Q^2}d}}{{4{\varepsilon _0}A}}\end{aligned}\)

Thus, the work must be done to reduce the separation of the plates from \(d\) to \(\frac{1}{2}d\) is \( - \frac{{{Q^2}d}}{{4{\varepsilon _0}A}}\).