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Question
A charge of `+2.0 xx 10^-8 C` is placed on the positive plate 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 "uF"` . Calculate the potential difference developed between the plates.
Solution
The charge on the positive plate is q1 and that on the negative plate is q2.
Given :
`q_1 = 2.0 xx 10^-8 C`
`q_2 = -1.0 xx 10^-8 C`
Now ,
Net charge on the capacitor = `((q_1 - q_2))/2 = 1.5 xx 10^-8 C`
The potential difference developed between the plates is given by `q = VC`
⇒ `V = (1.5 xx 10^-8)/(1.2 xx 10^-9)` = `12.5 V`
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