Question

A parallel-plate capacitor has capacitance 9.50 uF. (a) How much energy is stored in the capacitor if it is connected to a 7.00-V battery? (b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? (c) The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored? Need Help? Read It

Answer 1

Parallel plate capacitor having capacitance, C = 9.50 mu F V = 7 V Here, The energy stored, U = 1/2 Q . V Q = C V = 66.5 mu C So, Energy, U = 1/2 times 66.5 times 7 mu J rightarrow U = 232.75 mu J Capacitance, C = k A epsilon_0/d If distance is doubled, then, capacitance = 9.5/2 mu F = 4.75 mu F Then, Q = 66.5 mu c (will discharge) = 4.75 times V _of rightarrow V_4 = 14 volt Now, Energy, U = 1/2 times (66.5)^2/4.75 = 465.5 mu J \r\n Again Battery closed . v = 7 v so c = 4.75 mu F So, Energy stored, U = 1/2 times (4.75) (7) ^2 = 116.375 mu J