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Obtain the expression for capacitance for capacitance for a parallel plate capacitor . |
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Answer» Solution :Capacitance of a parallel plate capacitor : Consider a capacitor with two parallel plates each of cross - sectional area A and separated by a distance d . The electric FIELD between two INFINITE parallel plates is uniform and is given by `E = (sigma)/(epsilon_(0))` where `sigma ` is the surface charge density on the plates `sigma = (Q)/(A)` . If the separation distance d is very much smaller than the size of the plate `(d^(2) ltltA )` then the above result is USED even for finite -sized parallel plate capacitor . The electric field between the plates is `E= (Q)/(Aepsilon_(0))` Since the electric field is uniform the electric potential between the plates having separation d is given by `V = (Ed)= (Qd)/(Aepsilon_(0))` Therefore the capacitance of the capacitor is given by `C= (Q)/(V) = (Q)/((Qd)/(Aepsilon_(0)))= (epsilon_(0)A)/(d)`  From equation (3) it is evident that capacitance is DIRECTLY proportional to the area of cross section and is inversely proportional to the distance between the plates . This can be understood from the following . (i) If the area of cross - section of the capacitor plates is increased more charges can be distributed for the same potential difference . As a result the capacitance is increased. If the distance d between the two plates is reduced the potential difference between the plates (V= Ed) decreases with E constant . |
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