InterviewSolution
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InterviewSolution
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The plates of a parallel plate capacitor have an area of 90 cm^2? each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. View this energy as stored in the electrostatic field between the plates and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates. |
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Answer» Solution :Volume of the capacitor= A.d `= 90 xx 10^(-4) xx 2.5 xx 10^(-3) = 2.25 xx 10^(-3) = 2.25 xx 10^(-5) m^3` `:.` Energy per unit volume `u = (uu)/(Ad) = (2.55 xx 10^(-6))/(2.55 xx 10^(-5)) = 0.113 J m^(-3)` Relation between energy density u and electric FIELD E : We know that energy STORED in a capacitor is given by `uu = 1/2CV^2 = 1/2 (epsi_0AV^2)/d` and volume of the capacitor = Ad `:.` Energy density `u = u/(Ad)=((1/2epsi_0AV^2)/d)/(Ad) = 1/2epsi_0 V_2/d^2 = 1/2 epsi_0 E^2.` where `V/d = E` = electric field between the plates of capacitor. |
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