Question: (II) The electric field between the plates of a paper-separated \(\left( {{\bf{K = 3}}{\bf{.75}}} \right)\) capacitor is \({\bf{8}}{\bf{.24 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{V/m}}\). The plates are 1.95 mm apart, and the charge on each is \({\bf{0}}{\bf{.675}}\;{\bf{\mu C}}\). Determine the capacitance of this capacitor and the area of each plate.

The potential difference between the plates is the product of the electric field and the separation of the plates.

The expression for potential difference is given as:

\(V = Ed\)

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

The electric field inside the plates is,\(E = 8.24 \times {10^4}\;{\rm{V/m}}\).

The separation between the plates is,\(d = 1.95\;{\rm{mm}} = 1.95 \times {10^{ - 3}}\;{\rm{m}}\).

The charge in each plate is,\(Q = 0.675\,\mu {\rm{C}} = 0.675 \times {10^{ - 6}}\;{\rm{C}}\).

The dielectric constant of the material between the plates is, \(K = 3.75\).

The charge on the capacitor is given as:

\(\begin{aligned}{c}Q &= CV\\C &= \frac{Q}{V}\\C &= \frac{Q}{{Ed}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}C &= \frac{{0.675 \times {{10}^{ - 6}}\;{\rm{C}}}}{{\left( {8.24 \times {{10}^4}\;{\rm{V/m}}} \right) \times \left( {1.95 \times {{10}^{ - 3}}\;{\rm{m}}} \right)}}\\ &= 4.20 \times {10^{ - 9}}\;{\rm{F}}\end{aligned}\)

Thus, the capacitance of the capacitor is \(4.20 \times {10^{ - 9}}\;{\rm{F}}\).

The area of the capacitor is given as:

\(\begin{aligned}{c}C &= K{\varepsilon _0}\frac{A}{d}\\A &= \frac{{Cd}}{{K{\varepsilon _0}}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}A &= \frac{{\left( {4.20 \times {{10}^{ - 9}}\;{\rm{F}}} \right)\left( {1.95 \times {{10}^{ - 3}}\;{\rm{m}}} \right)}}{{3.75 \times \left( {8.854 \times {{10}^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right)}}\\A &= 0.247\;{{\rm{m}}^{\rm{2}}}\end{aligned}\)

Thus, the area of the plates is \(0.247\;{{\rm{m}}^{\rm{2}}}\).