A parallel plate capacitor of capacity `C_(0)` is charged to a potential `V_(0), E_(1)` is the energy stored in the capacitor when the battery is disconnected and the plate separation is doubled, and `E_(2)` is the energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is dounled. find the ratio `E_(1)//E_(2)`.

Options

A. 2
B. `(3)/(2)`
C. 2
D. `(1)/(2)`

Answer

A

Explanation

Charge on the capacitor is
`Q_(0) = C_(0)V_(0)` …(i)
In first case, When the battery is disconnected, the charge remains the same. Therefore energy stored in the capacitor is
`U_(1) = (1)/(2) (Q_(0)^(2))/(C)` where `C = (C_(0))/(2)`
`:. U_(1) = (Q_(0)^(2))/(C_(0)) = (C_(0)^(2)V_(0)^(2))/(C_(0))` (Using (i))
or `U_(1) = C_(0)V_(0)^(2)`
In second case, When the battery is kept connected, the potential remains the same. Therefore energy stored in the capacitor is
`U_(2) = (1)/(2)CV_(0)^(2) = (1)/(2)((C_(0))/(2)) V_(0)^(2) = (1)/(4)C_(0)V_(0)^(2)` ...(iii)
Divide (ii) by (iii), we get `(U_(1))/(U_(2)) = (C_(0)V_(0)^(2))/((1)/(4) C_(0)V_(0)^(2)) = 4`