(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 $3.00 \times 10^{3}$ is applied to it?
Strategy
Finding the capacitance C is a straightforward application of the equation $C=\varepsilon_{0} A / d$ . Once C is found, the charge stored can be found using the equation Q = CV .

Question

(a) What is the capacitance of a parallel plate capacitor with metal plates, each of area [latex]1.00 m ^{2}[/latex] , separated by 1.00 mm? (b) What charge is stored in this capacitor if a voltage of [latex]3.00 imes 10^{3}[/latex] is applied to it?
Strategy
Finding the capacitance C is a straightforward application of the equation [latex]C=\varepsilon_{0} A / d[/latex]. Once C is found, the charge stored can be found using the equation Q = CV .

Accepted Answer

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

[latex]C=\varepsilon_{0} \frac{A}{d}=\left(8.85 imes 10^{-12} \frac{ F }{ m } ight) \frac{1.00 m ^{2}}{1.00 imes 10^{-3} m }[/latex] (19.54)

[latex]=8.85 imes 10^{-9} F =8.85 nF[/latex].

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

[latex]Q=C V=\left(8.85 imes 10^{-9} F ight)\left(3.00 imes 10^{3} V ight)[/latex] (19.55)

= 26.6 μC.

Discussion for (b)
This charge is only slightly greater than those found in typical static electricity. Since air breaks down at about [latex]3.00 imes 10^{6}[/latex] , more charge cannot be stored on this capacitor by increasing the voltage.

Question 19.8: (a) What is the capacitance of a parallel plate capacitor wi...

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