A parallel plate capacitor having common plate area `50" cm"^(2)` and plate separation `2.0` mm is connected to a battery of `6.0` V. Calculate the charge on the capacitor and the work done by the battery.

Explanation

As we know that `Q=CV` . . .(i)
For a parallel plate capacitor,
`C=(epsilon_(0)A)/(d)` . . . (ii)
where, A = area of the plate
and d = distance between plate.
From Eqs. (i) and (ii), we get
`Q=((epsilon_(0)A)/(d))V`
`=(8.85xx10^(-12)xx50xx10^(-4))/(2.0xx10^(-3))xx6.0`
`rArr" "Q=1.33xx10^(-10)C`
`:.` Work done by the battery,
`W=QV`
`=1.33xx10^(-10)xx6`
`:." "W=7.98xx10^(-10)J`