Question

Consider a parallel-plate capacitor whose plates have area 0.00450 m2 and separation

distance 0.000750 m.

1. Calculate the capacitance of this capacitor.

2. The capacitor is connected to a 12.0-V power source.

a. How much charge is stored in the capacitor plates?

b. What is the magnitude of the electric field produced between the plates of

the capacitor by the stored charge?

3. We want to construct a cylindrical capacitor with an equivalent capacitance. If

we are to use the same separation distance for the cylinders, then we can rewrite

the capacitance of the cylindrical capacitor to be

C =

2πε0L

ln d

.

a. How long should our cylindrical capacitor be if it is to have the same

capacitance as the parallel-plate capacitor?

Answer 1

1) We can find the capacitance of the parallel-plate capacitor as follows:

$$C=\epsilon_0\dfrac{A}{d},$$ $$C=8.85\cdot10^{-12}\ \dfrac{F}{m}\cdot\dfrac{4.5\cdot10^{-3}\ m^2}{7.5\cdot10^{-4}\ m}=53.1\cdot10^{-12}\ F=53.1\ pF.$$

2) a) We can find the charge stored in the capacitor plates as follows:

$$C=\dfrac{Q}{\Delta V},$$ $$Q=C\Delta V,$$ $$Q=53.1\cdot10^{-12}\ F\cdot12\ V=637\ pC.$$

b) We can find the magnitude of the electric field produced between the plates of

the capacitor by the stored charge as follows:

$$E=\dfrac{V}{d}=\dfrac{12\ V}{7.5\cdot10^{-4}\ m}=1.6\cdot10^4\ V=16\ kV.$$

3)

$$C=\dfrac{2\pi\epsilon_0L}{ln\ \!d},$$ $$L=\dfrac{Cln\ \!d}{2\pi\epsilon_0},$$ $$L=\dfrac{53.1\cdot10^{-12}\ F\cdot ln\ \!7.5\cdot10^{-4}\ m}{2\pi\cdot8.85\cdot10^{-12}\ \dfrac{F}{m}}=6.9\ m.$$