(II) The electric field between two parallel square metal plates is 130 N/C. The plates are 0.85 m on a side and are separated by 3.0 cm. What is the charge on each plate (assume equal and opposite)? Neglect edge effects.

Gauss’s law gives a relation to determine the electric flux passing through a closed surface. According to Gauss’s law, the total electric flux\(\left( {{{\bf{\Phi }}_{\bf{E}}}} \right)\)over a closed surface placed in a vacuum is\(\frac{{\bf{1}}}{{{{\bf{\varepsilon }}_{\bf{0}}}}}\)times the total charge contained in it.

The expression for total electric flux is,

\({\Phi _E} = \frac{Q}{{{\varepsilon _0}}}\)

Here, Q is the charge enclosed in the surface and \({\varepsilon _0}\) is the absolute electrical permittivity of free space whose value is \(8.85 \times {10^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}\).

The electric field between two parallel metal plates is, \(E = 130\;{\rm{N/C}}\)

The length of the side of the metal plate is, \(l = 0.85\;{\rm{m}}\)

The separation between two parallel metal plates is,\(d = 3.0\;{\rm{cm}} = 3.0 \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}\)

The area of each square metal plate is,

\(\begin{aligned}{c}A = {l^2}\\ = {\left( {0.85\;{\rm{m}}} \right)^2}\\ = 72.25 \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{{\rm{m}}^2}\end{aligned}\)

The charge on two metal plates is equal and opposite. Let the magnitude of the charge on each plate be Q. Since the electric field is perpendicular to the metal plate; thus the angle between the electric field and normally drawn perpendicularly outwards to the metal plate is zero.

The net electric flux through the metal plate is:

\(\begin{aligned}{c}{\Phi _E} = EA\cos \theta \\ = \left( {130\;{\rm{N/C}}} \right) \times \left( {72.25 \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{{\rm{m}}^2}} \right)\cos 0^\circ \\ = 93.9\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/C}}\end{aligned}\)

According to Gauss’s law, the total electric flux over the closed metal surface is given as:

\({\Phi _E} = \frac{Q}{{{\varepsilon _0}}}\)

The charge on each plate is:

\(\begin{aligned}{c}Q = {\varepsilon _0}{\Phi _E}\\ = \left( {8.85 \times {{10}^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {93.9\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/C}}} \right)\\ = 831.0 \times {10^{ - 12}}\;{\rm{C}}\\ = 8.3 \times {10^{ - 10}}\;{\rm{C}}\end{aligned}\)

Thus, the magnitude of the charge on each plate is \(8.3 \times {10^{ - 10}}\;{\rm{C}}\).