Example 19.8 from College Physics, 19.5 Capacitors and Dielectrics

Capacitance and Charge Stored in a Parallel Plate Capacitor

(a) What is the capacitance of a parallel plate capacitor with metal plates, each of area $1.00 m^{2}$ , separated by 1.00 mm? (b) What charge is stored in this capacitor if a voltage of $3.00 \times 10^{3} V$ is applied to it?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Finding the capacitance $C$ is a straightforward application of the equation $C = ε_{0} A / d$ . Once $C$ is found, the charge stored can be found using the equation $Q = CV$ .
Solution for (a)

Entering the given values into the equation for the capacitance of a parallel plate capacitor yields

$\begin{array}{l} C & = & ε_{0} \frac{A}{d} = (8.85 \times 10^{–12} \frac{F}{m}) \frac{1.00 m^{2}}{1.00 \times 10^{–3} m} \\ = & 8.85 \times 10^{–9} F = 8.85 nF. \end{array}$
19.54
Discussion for (a)

This small value for the capacitance indicates how difficult it is to make a device with a large capacitance. Special techniques help, such as using very large area thin foils placed close together.
Solution for (b)

The charge stored in any capacitor is given by the equation $Q = CV$ . Entering the known values into this equation gives
$\begin{array}{l} Q & = & CV = (8.85 \times 10^{–9} F) (3.00 \times 10^{3} V) \\ = & 26.6 μC. \end{array}$
19.55
Discussion for (b)

This charge is only slightly greater than those found in typical static electricity. Since air breaks down at about $3 . 00 \times 10^{6} V/m$ , more charge cannot be stored on this capacitor by increasing the voltage.

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