A charge of` +2.0xx 10 ^-8 C` is placed on the positive plate and a charge of `-1.0xx10 ^-8 C `at negative plate. And capacitance of capacitor is `1.2 xx (10^-3) mu F` . Calculate the potential difference developed between the plates.

Explanation

`q_1 = 2.0 xx 10^(-8) C`,
`q_2 = - 1.0 xx 10^(-8) C `
Net charge `= (q_1 - q_2) / 2 `
`= 1.5 xx 10^(-8) C `
`q = VC`
`V = 1.5 xx 10^(-8) / 1.2 xx 10^(-9) = 12.5 V`