If the distance between plates of a capacitor having capacity C and charge Q is doubled then the work done will be : (a)Q^2/4C (b)Q^2/2C
Answer (1)
11th Jan, 2019
Hello,
The energy initially stored by the capacitor is (1/2 C*V^2) or( Q^2 / 2*C)
When the plate separation is doubled, the charge remains constant and the capacitance is halved. The work done in separating the plates is the difference in energy stored:this is
Q^2 / 2*0.5*C - Q^2 / 2*CT
This represents double the initial energy less the initial energy. The work done in adding this energy is Q^2 / 2*C.that is answer is option B
Hope this helps......
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