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

Options

A. 4
B. `3/2`
C. 2
D. `1/2`

Answer

(a)

Explanation

Capacitance of parallel plate capacitor ,`C_(0) = epsilon_(0)A/d`
Charge stored in the capaciyor ,Q = `C_(0)V_(0)`
When battery is disconnected , energy,
`E_(1) = 1/2 Q^(2)/C = 1/2 Q^(2)/2C_(0)/2 = Q^(2)/C_(0)`
When battery is disconnected energy,
`E_(1) = 1/2Q^(2)/C = 1/2 Q^(2)/C_(0)/2 = Q^(2)/C_(0)`
When battery is connected ,
Energy, `E_(2) = 1/2 CV_(0)^(2) = 1/2 C_(0)/2 V_(0)^(2)= 1/4 (C_(0)V_(0)^(2))`
`E_(1)/E_(2) = C_(0)V_(0)^(2)/1/4C_(0)V_(0)^(2) = 4/ 1`
` :. E_(1)/E_(2) = 4`