Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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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. - Physics

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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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Solution 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. Concept: Electric Potential Difference.
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