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Next: Example 6.2: Dielectric filled Up: Capacitance Previous: Worked Examples

Example 6.1: Parallel plate capacitor

Question: A parallel plate capacitor consists of two metal plates, each of area $A=150\,{\rm cm}^2$, separated by a vacuum gap $d=0.60$ cm thick. What is the capacitance of this device? What potential difference must be applied between the plates if the capacitor is to hold a charge of magnitude $Q=1.00\times 10^{-3}\,\mu{\rm C}$ on each plate?
Solution: Making use of formula (108), the capacitance $C$ is given by

C= \frac{(8.85\times 10^{-12})\,(150\times 10^{-4})}{(0.6\times 10^{-2})}
=2.21\times 10^{-11} = 22.1\,{\rm pF}.

The voltage difference $V$ between the plates and the magnitude of the charge $Q$ stored on each plate are related via $C=Q/V$, or $V = Q/C$. Hence, if $Q=1.00\times 10^{-3}\,\mu{\rm C}$ then

V = \frac{ (1.00\times 10^{-9})}{(2.21\times 10^{-11})} = 45.2\, {\rm V}.

Richard Fitzpatrick 2007-07-14