An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its pl

Question

An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.
a) What is the energy stored in the capacitor?

Now a conductor is inserted into the capacitor. The thickness of the conductor is 1/3 the distance between the plates of the capacitor and is centered inbetween the plates of the capacitor.
b) What is the charge on the plates of the capacitor?

c) What is the capacitance of the capacitor with the conductor in place?

d) What is the energy stored in the capacitor with the conductor in place?

Explanation / Answer

Given Capacitance of the capacitor , C = 1F Charge on the plates is ,Q = 45C a) Energy stored in the capacitor is                          W = Q^2/2C                             = (45 *10^-6 C)^2 / 2*(1 *10^-6 F)                              = 1.01*10^-3 J --------------------------------------------------------------- The thickness of the conductor is , t = (1/3) *d b) The charge on the plates of the capacitor remains      unchanged ,that is Q = 45C c) Electric field is non zero over (1-1/3) = 2/3 of d     From the relation V = Ed,     The voltage has changed by a factor of 2/3     The capacitance C' = Q/V                               C' = (3/2) C = (3/2) * 1F                                C' = 1.5 F d)Energy stored is W' = Q^2/2C'                             = (45 *10^-6 C)^2 / 2*(1. 5*10^-6 F)                              = 6.75*10^-4 J        --------------------------------------------------------------- The thickness of the conductor is , t = (1/3) *d b) The charge on the plates of the capacitor remains      unchanged ,that is Q = 45C c) Electric field is non zero over (1-1/3) = 2/3 of d     From the relation V = Ed,     The voltage has changed by a factor of 2/3     The capacitance C' = Q/V                               C' = (3/2) C = (3/2) * 1F                                C' = 1.5 F d)Energy stored is W' = Q^2/2C'                             = (45 *10^-6 C)^2 / 2*(1. 5*10^-6 F)                              = 6.75*10^-4 J                             = (45 *10^-6 C)^2 / 2*(1. 5*10^-6 F)                              = 6.75*10^-4 J       

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