A parallel plate capacitor having area `25 cm^2` and separation 1.00 mm is Connected to a battery of 6.0 V. Calculate the charge flown through the battery. How much work has been done by the battery during this process? `(Given varepsilon_0=8.85xx10^-12C^2N^-1m^-2)`

Explanation

`A = 25 cm^(2) = 2.5 xx 10^(-4) m^(2)`
`d = 1.0 mm = 1.0 xx 10^(-3) m`
V = 6 volt
`epsilon_(0)= 8.854 xx 10^(-12) C^(2) N^(-1) m^(-2)`
Charge `q = CV`
`q =(epsilon_(0)A)/dV`
`=(8.854 xx 10^(-12) C^(2) N^(-1) m^(-2)`
Charge q=CV
`q=(epsilon_(0)A)/dV`
`=8.854 xx 10^(-12) xx 25 xx 10^(-4))/10^(-3) xx 6`
`q = 1.33 xx 10^(-10) C`
Work done by battery `=1/2 qV`
`=1/2 xx 1.33 xx 10^(-10) xx 6`
`=3.99 xx 10^(-10) J`