Sum

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.

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#### Solution

The charge on the positive plate is q_{1}_{ }and that on the negative plate is q_{2}.

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`

Concept: Electric Potential Difference

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