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Question
Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.
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Solution
The capacitance of a capacitor is the amount of charge which creates unit potential difference between collecting plate and condensing plate after giving charge on the collecting plate.
Parallel Plate Capacitor
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A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
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Let A be the area of each plate and d the separation between them. The two plates have charges Q and −Q.
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Surface charge density of plate 1, σ = Q/A, and that of plate 2 is σ.
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Electric field in different regions:
Outer region I,
`E = σ/(2ε_0) - σ/(2ε_0 ) = 0`
In the inner region between plates 1 and 2, the electric fields due to the two charged plates add up. So,
\[E = \frac{\sigma}{2 \epsilon_0} + \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}\]
E=σ2ε0+σ2ε0=σε0=Qε0
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The direction of electric field is from the positive to the negative plate.
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For uniform electric field, potential difference is simply the electric field multiplied by the distance between the plates, i.e.
`V=Ed = 1/c (Qd)/A`
Capacitance C of the parallel plate capacitor in vacuum is
`C = Q/V =(ε_0A)/d `
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