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A 4 (mu)F capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2(mu)F capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation.
first we find the charge of the first capacitor by Q=CV,
where V=200v and C=4(mu)f,
then, we find the carge on the second capacitor,
as the final voltage is same Q1/C1=Q2/C2;
the we find the energy initially of the first capacitor and the the final energies of the first and second capacitors. E=Q2/2C
we subtract the initial from the final energy, we get the ans
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