A parallel-plate capacitor has plates of area `A` and separation `d` and is charged to a potential difference `V`. The charging battery is then disconnected, and the plates are pulled apart until their separation is `2d`. Derive expression in terms of `A, d` and `V` for
(a) the new potential difference
(b) the initial and final stored energies, `U_i` and `U_f` and
(c) the work required to increase the separation of plates from d to `2d`.

Options

A. `A,B,C`
B. `A,C,D`
C. `A,B,D`
D. `A,B,C,D`

Answer

D

Explanation

Charge remains same, `Q=CV=(Ain_(0)V)/(d)`
`Q=CV=(c)/(2)V' impliesV'=2V,(A)` is O.K.
`(U_(1))/(U_(2))=(Q^(2)//2C)/(Q^(2)//2.(C)/(2))=(1)/(2),(B)` is O.K.
`W_(1rarr2)=U_(2)-U_(1)=(Q^(2))/(2.(c)/(2))-(Q^(2))/(2C)=(Q^(2))/(2C)=(1)/(2)CV^(2)`
`=(1)/(2)(Ain_(0)V^(2))/(d),(c)` is O.K.
Energy density `u=(1)/(2)in_(0)E^(2)=(1)/(2)in_(0)((V)/(d))^(2)`
`u'=(1)/(2)in_(0)E'^(2)=(1)/(2)in_(0)((2V)/(2d))^(2)`
`u=u',(D)` is O.K.