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Deduce an expression for the capacitance ofa parallel plate capacitor having plate area 'A' and plate separation 'd'. |
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Answer» Solution :Consider a parallel PLATE capacitor consisting of TWO plates each of surfaces area A , held parallel to each other in AIR medium such that the separation between the plates is d . Charges on the two plates are `+Q` and `-Q` , respectively . Then surface density of charge on either plate `|sigma| = (Q)/(A)` Electric field between the plates of capacitor is uniform every where , having a value `E = E_(1) + E_(2) = (sigma)/(2 in_(0)) + (sigma )/(2 in_(0)) = (sigma)/(in_(0))` The electric field E is directed from +ve plate to the -ve plate . If V be the potential difference between the plates , then `V = E d = (sigma)/(in_(0)) d = (Qd)/(in_(0) A)` `therefore` Capacitance of the parallel plate capacitor `C = (Q)/(V) = (in_(0) A)/(d)` If a dielectric medium of dielectric CONSTANT .K. is completely filled between the two plates of capacitor , then electric field `E = (sigma)/(K in_(0)) = (Q)/(K A in_(0))` In that case `V = Ed = (Qd)/(K A in_(0)) implies` Capacitance `C = (Q)/(V) = (K in_(0) A)/(d)` 
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