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The plates of a parallel plate capacitor have an area of 100 cm^(2) each and are separated by 3mm. The capacitor is charged by connecting it to a 400 V supply. (a) Calculate the electrostatic energy stored in the capacitor. (b) If a dielectric of dielectric constant 2.5 is introduced between the plates of the capacitor then find the electrostatic energy stored and also change in the energy stored. |
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Answer» Solution :Given `A = 100 CM^(2) = 100 xx 10^(-4) m^(2), d = 3mm = 3 xx 10^(-3)m, V = 400 V, u_(1) = ?, K = 2.5, u_(2) = ?` ALSO `u_(2) - u_(1) = ?` We have `U = (1)/(2) C V^(2)` `C = (epsilon_(0)A)/(d) = (8.854 xx 10^(-12) xx 100 xx 10^(-4))/(3 xx 10^(-3)) = (8.854 xx 10^(-16))/(3 xx 10^(-3)) = 295.13 xx 10^(-13)` `C = 29.513 xx 10^(-12) F` (i) `U_(1) = (1)/(2)CV^(2)` `= (1)/(2) 29.513 xx 10^(-12) xx (400)^(2) = (4722080)/(2) xx 10^(-12) = 2361040 xx 10^(-12)` `U_(1) = 23.610 xx 10^(-7) J` (ii) If dielectric medium K = 2.5 is introduced between the PLATES of the capacitors, the its capacitance is `C = (epsilon_(0) KA)/(d) = (8.854 xx 10^(-12) xx 2.5 xx 100 xx 10^(-4))/(3 xx 10^(-3)) = (2213.5)/(3 xx 10^(-3)) xx 10^(16) = 737.833 xx 10^(-13)` `C = 73.7833 xx 10^(-12) F` `U_(2) = (1)/(2) C V^(2) = (1)/(2) 73.7833 xx 10^(-12) xx (400)^(2) = 5902661 xx 10^(-12)` `U_(2) = 59.02664 xx 10^(-7)J` CHANGE in energy `Delta U = U_(2) - U_(1) = 59.02664 xx 10^(-7) - 23.610 xx 10^(-7)` `Delta U = 35.14661 xx 10^(-7)J` |
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