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Magazine Chapter 6: Q82GP (page 138)
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 鈥渉omemade鈥� capacitor by placing a sheet of \({\bf{21 \times 14}}\;{\bf{cm}}\) paper between two aluminum foil sheets (Fig. 17鈥�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鈥�48 Problem 82.
Short Answer
(a) The capacitance of the device is \(8.75 \times {10^{ - 9}}\;{\rm{F}}\).
(b) The amount of charge stored in the capacitor is \(14.43 \times {10^{ - 6}}\;{\rm{C}}\).
Step by step solution
Variables on which the capacitance value depends
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.
Given information
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, \(A = \left( {21 \times 14} \right)\;{\rm{c}}{{\rm{m}}^{\rm{2}}}\).
(a) Evaluation of the capacitance of the device
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 \(8.75 \times {10^{ - 9}}\;{\rm{F}}\).
(b) Evaluation of the amount of charge stored in the capacitor
The amount of charge stored in the capacitor can be calculated as:
Thus, the amount of charge stored in the capacitor is \(14.43 \times {10^{ - 6}}\;{\rm{C}}\).
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