Question: Paper has a dielectric constant K = 3.7 and a dielectric strength of \({\bf{15 \times 1}}{{\bf{0}}^{\bf{6}}}\) V/m. Suppose that a typical sheet of paper has a thickness of 0.11 mm. You make a “homemade” capacitor by placing a sheet of \({\bf{21 \times 14}}\;{\bf{cm}}\) paper between two aluminum foil sheets (Fig. 17–48) of the same size. (a) What is the capacitance C of your device? (b) About how much charge could you store on your capacitor before it would break down?

FIGURE 17–48 Problem 82.

The capacitance value relies on the value of the charge and the value of the potential difference. Its value is altered inversely to the value of the potential difference.

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

The dielectric strength is, \(E = 15 \times {10^6}\) V/m.

The thickness of the paper is, \(d = 0.11\;{\rm{mm}}\).

The area of the paper is, \(A = \left( {21 \times 14} \right)\;{\rm{c}}{{\rm{m}}^{\rm{2}}}\).

The capacitance of the device can be calculated as:

\(C = \frac{{k{\varepsilon _o}A}}{d}\)

Here, \({\varepsilon _o}\) is the permittivity of free space.

Substitute the values in the above expression.

Thus, the capacitance of the device is \(8.75 \times {10^{ - 9}}\;{\rm{F}}\).

The amount of charge stored in the capacitor can be calculated as:

Thus, the amount of charge stored in the capacitor is \(14.43 \times {10^{ - 6}}\;{\rm{C}}\).