Question

(II) If a capacitor has opposite \({\bf{4}}{\bf{.2}}\;{\bf{\mu C}}\) charges on the plates, and an electric field of \({\bf{2}}{\bf{.0}}\;{\bf{kV/mm}}\) is desired between the plates, what must each plate’s area be?

Short answer

The area of each plate is\(0.24\;{{\rm{m}}^2}\).

Step-by-step solution

Understanding of electric field

The electric field between the parallel plates is equivalent to the ratio of voltage supplied to the separation distance between the plates.

The electric field between the plates is given by,

\(E = \frac{V}{d}\) …… (i)

Here, E is the electric field, V is the potential difference and d is the separation between the plates.

Given Data

The electric field between the plates is,\(E = 2.0\;{\rm{kV/mm}}\).

The charge on each plate is,\(Q = 4.2\;\mu {\rm{C}}\).

Determination of the area of each plate

The charge on each plate of the capacitor is given by,

\(\begin{aligned}Q &= CV\\Q &= \left( {\frac{{{\varepsilon _0}A}}{d}} \right)V\end{aligned}\)

Use equation (i) in the above expression.

\(\begin{aligned}Q &= {\varepsilon _0}AE\\A &= \frac{Q}{{{\varepsilon _0}E}}\end{aligned}\)

Here, \({\varepsilon _0}\) is the permittivity of free space, Ais the area, Q is the charge and E is the electric field.

Substitute the values in the above expression.

\(\begin{aligned}A &= \frac{{4.2\;\mu {\rm{C}} \times \frac{{{{10}^{ - 6}}\;{\rm{C}}}}{{1\;\mu {\rm{C}}}}}}{{\left( {8.85 \times {{10}^{ - 12}}\;{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}} \right)\left( {2\;{\rm{kV/mm}} \times \frac{{{{10}^6}\;{\rm{V/m}}}}{{1\;{\rm{kV/mm}}}}} \right)}}\\A &\approx 0.24\;{{\rm{m}}^2}\end{aligned}\)

Thus, the area of each plate is \(0.24\;{{\rm{m}}^2}\).