A pararllel plate capacitor has plates of area `A` and separation `d` and is charged to potential diference `V`. The charging battery is then disconnected and the plates are pulle apart until their separation is `2d`. What is the work required to separate the plates?

Options

A. `2epsilon_(0)AV^(2)//d`
B. `epsilonAV^(2)//d`
C. `3epsilon_(0)AV^(2)//2d`
D. `epsilon_(0)AV^(2)//2d`

Answer

D

Explanation

`W=U_(2)-U_(1)=q^(2)/2[1/C_(2)-1/C_(1)]`
`C_(1)=(epsilon_(0)A)/d, C_(2)=C_(1)/2=(epsilonA)/(2d)`
`q=C_(1)V=(epsilon_(0)AV)/d`
Solve to get `W=1/2(epsilon_(0)AV^(2))/d`
Alternatively:
`W=Fd=Q^(2)/(2Aepsilon_(0))d=(C_(1)^(2)V^(2))/(2epsilon_(0)A)d=1/2(epsilon_(0)AV^(2))/d`.