A parallel plate capacitor is charged by connecting it to a battery. The battery is disconnected and the plates of the capacitor are pulled apart to make the separation between the plates twice. Again the capacitor is connected to the battery (with same polarity) then the potential difference between the plates increases when the plates are pulled apart (yes or not)

Options

A. `1/2 CV^2`
B. `3/2 CV^2`
C. `-3/2 CV^2`
D. `CV^2`

Answer

A

Explanation

Capacitance of a parallel plate capacitor is
`C=(epsilon_0A)/d " "....(i)`
Where A is the area of each plate and d is the distance between the plates.
Initial energy stored in the capacitor,
`U_i =1/2CV^2 " "..(ii)`
When the separation between the plates is doubled, its capacitance becomes
`C.=(epsilon_0A)/(2d) =1/2 (epsilon_0A)/(d) =C/2 ` [Using (i)] `" "....(iii)`
As the battery is disconnected, so charged capacitor becomes isolated and charge on it will remain constant,
`:. Q. =Q`
C.V. = CV [As A =CV]
`V.=(C/(C.))V=C/((C/2))V=2V " "...(iv)`
Final energy stored in the capacitor
`U_f =1/2C.V.^(2) =1/2 (C/2)(2V)^2 =CV^2 " "...(v)`
[Using (iii) and (iv)]
Work done, `W=U_f-U_i=CV^2-1/2CV^2=1/2CV^2`
Aliter : Without dielectric -
Charge (Q) = CV
Initial energy `(U_i)=1/2 Q^2/C=1/2CV^2`
When battery is disconnected and d. = 2d
`C.=(Aepsilon_0)/(d.)=C/2`
As charge remains constant –
Final energy `(U_f)=Q^2/(2C.)=((CV)^(2))/(2.C/2)=2U_i`
Work done `W=U_f -U_i=2U_i-U_i=U_i=1/2CV^2`