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Question
Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
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Solution
A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
Let A be the area of each plate and d be the separation between them. The two plates have charges Q and −Q. Plate 1 has surface charge density, σ = Q/A,
And plate 2 has a surface charge density −σ.
Electric field in different regions:
Outer region I:
`E= sigma/(2epsilon_0)-sigma/(2epsilon_0)=0`
Outer region II:
`E=sigma/(2epsilon_0)-sigma/(2epsilon_0)=0`
In the inner region between plates 1 and 2, the electric fields due to the two charged plates add up.
This gives:
`E=sigma/(2epsilon_0)+sigma/(2epsilon_0)=sigma/epsilon_0=Q/epsilon_0=Q/(epsilon_0A)`
The direction of electric field is from positive to the negative plate. For uniform electric field, potential difference is simply the electric field times the distance between the plates.
`V=E""d=1/epsilon_0(Qd)/A`
Capacitance (C) of the parallel plate capacitor, `C=Q/V=(epsilon_0A)/d`
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