Question: A capacitor is made from two 1.1-cm-diameter coins separated by a 0.10-mm-thick piece of paper \(\left( {{\bf{K = 3}}{\bf{.7}}} \right)\). A 12-V battery is connected to the capacitor. How much charge is on each coin?

The capacitance value can be obtained by dividing the value of the charge by the value of the potential difference. Its value is directly related to the value of the charge.

The diameter of the coin is, \(d = 1.1\;{\rm{cm}}\)

The thickness of the paper is, \(x = 0.10\;{\rm{mm}}\)

The dielectric constant is, \(K = 3.7\)

The voltage of the battery is, \(V = 12\;{\rm{V}}\)

The area of the coin can be calculated as:

\(\begin{aligned}{c}A &= \frac{\pi }{4}{d^2}\\ &= \frac{\pi }{4}{\left( {\left( {{\rm{1}}{\rm{.1}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right]^2}\\ &= 9.5 \times {10^{ - 5}}\;{{\rm{m}}^{\rm{2}}}\end{aligned}\)

The expression for the capacitance is as follows:

\(C = \frac{{k{\varepsilon _{\rm{o}}}A}}{x}\) … (i)

The relationship between the capacitance, charge, and voltage is as follows:

\(C = \frac{Q}{V}\)

Substitute the value of equation (i) in the above equation.

\(\begin{aligned}{c}\frac{{k{\varepsilon _{\rm{o}}}A}}{x} &= \frac{Q}{V}\\Q &= \frac{{k{\varepsilon _{\rm{o}}}AV}}{x}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}Q = \frac{{\left( {3.7} \right)\left( {9 \times {{10}^{ - 12}}\;{{{{\rm{C}}^{\rm{2}}}} \{\left/ {{{{\rm{C}}^{\rm{2}}} {{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}}}} \right. {{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}}}} \right)\left( {9.5 \times {{10}^{ - 5}}\;{{\rm{m}}^{\rm{2}}}} \right)\left( {12\;{\rm{V}}} \right)}}{{\left( {0.10\;{\rm{mm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{mm}}}}} \right)}}\\ = 3.7 \times {10^{ - 10}}\;{\rm{C}}\end{aligned}\)

Thus, the charge on each coin is \(3.7 \times {10^{ - 10}}\;{\rm{C}}\).