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Question
A charge of 20 µC is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10 µF. Calculate the potential difference developed between the plates.
Solution
Given :
Capacitance of the isolated capacitor = 10 µF
Charge on the positive plate = 20 µC
Effective charge on the capacitor = `(20-0)/2 = 10 "uC"`
The potential difference between the plates of the capacitor is given by `V = Q/C`
`therefore "Potential difference" = (10 "uC")/(10 "uF")`= `1 "V"`
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