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A charge of +2.0xx 10^(-8) C is placed o...

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.

Text Solution

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Here, `q_(1) = 2.0xx10^(-8) C`.
`q_(2) = 1.0xx10^(-8) C`
`C = 1.2xx10^(-3) muF = 1.2xx10^(-9) F, V = ?`
`V = (q_(1) - q_(2))/(2C) = (2.0xx10^(-8) -(-1.0xx10^(-8)))/(2xx1.2xx10^(-9))`
12.5V
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