A charge of` +2.0xx 10 ^-8 C` is placed on the positive place and a charge of -`1.0 xx 10 ^-8 C` on the negative plate of a parallel- plate capacitor of capacitance ` 1.2 xx (10^-3) mu` F. Calculate the potential difference developed between the plates.

Explanation

Let `sigma_(1)` and `sigma_(2)` be charges on the two plates and E be electric field between the plates due to these charge.
Then, `E = (sigma_(1) - sigma_(2))/(2 in_(0)) = ((q_(1))/(A) - (q_(2))/(B))/(2 in_(0)) = (q_(1) -q_(2))/(2 A in_(0))`
P.d between plates `V = Ed = (q_(1) - q_(2))/(2A in_(0)) .d`
`= (q_(1) -q_(2))/(2.(in_(0)A)/(d)) = (q_(1) - q_(2))/(2C) = ((2 xx 10^(-3))- (-1xx 10^(-3)))/(2 xx 1.2 xx 10^(-6))`
= 1250 Volt