Question:Dry air will break down if the electric field exceeds \({\bf{3}}{\bf{.0 \times 1}}{{\bf{0}}^{\bf{6}}}\;{\bf{V/m}}\). What amount of charge can be placed on a parallel-plate capacitor if the area of each plate is \({\bf{65}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\)?

The capacitance of a capacitor relies on the area of each plate and the separation between the two plates.

The capacitance of a capacitor is given as follows:

\(C = \frac{Q}{V} = {\varepsilon _0}\frac{A}{d}\) … (i)

Here, Q is the charge;V is the voltage applied across the plates;A is the area of each plate;d is the separation between the plates; \({\varepsilon _0}\)is the absolute electrical permittivity of the free space whose value is \(8.85 \times {10^{ - 12}}\;{{\rm{C}}^{\rm{2}}}\;{{\rm{N}}^{ - 1}}\;{{\rm{m}}^{ - 2}}\).

The electric field between the plates of the capacitor is\(E = 3.0 \times {10^6}\;{\rm{V/m}}\).

The area of each plate is \(A = 65\;{\rm{c}}{{\rm{m}}^{\rm{2}}} = 65 \times {10^{ - 4}}\;{{\rm{m}}^{\rm{2}}}\).

The magnitude of the electric field is related to the potential difference (V) across the plates of the capacitor and plate separation (d) as follows:

\(E = \frac{V}{d}\) … (ii)

From equations (i) and (ii), the charge on the capacitor plates is as follows:

\(\begin{array}{c}Q = CV\\ = \left( {{\varepsilon _0}\frac{A}{d}} \right)V\\ = {\varepsilon _0}A\left( {\frac{V}{d}} \right)\\ = {\varepsilon _0}AE\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}Q = \left( {8.85 \times {{10}^{ - 12}}\;{{\rm{C}}^{\rm{2}}}\;{{\rm{N}}^{ - 1}}\;{{\rm{m}}^{ - 2}}} \right)\left( {65 \times {{10}^{ - 4}}\;{{\rm{m}}^{\rm{2}}}} \right)\left( {3.0 \times {{10}^6}\;{\rm{V/m}}} \right)\\ = 1725.75 \times {10^{ - 10}}\;{\rm{C}}\\ = 1.{\rm{7}} \times {10^{ - 7}}\;{\rm{C}}\end{array}\)

Thus, the charge that can be placed on the parallel-plate capacitor is \(1.{\rm{7}} \times {10^{ - 7}}\;{\rm{C}}\).