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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


Kusum yadav 11th Jan, 2019
Answer (1)
Akhila Nagasree 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...... 

1 Comment
Comments (1)
11th Jan, 2019
Yeah...i got it...thank u 
Reply

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