Question

A capacitor has a capacitance of 7.28 muF. What amount of charge must be placed on each of its plates to make the potential difference between its plates equAl to 25.0 V?

Answer 1

We know that charge on capacitor is Q=CV=7.28×25=182μC

The charges on each plate of capacitor are equal and opposite. One plate will get charge −Q and the other plate is +Q.......✌✌

Answer 2

Given:-  

Capacitance C = 7.28μF.

Voltage V = 25.0 Volt.

To Find :-

Amount of charge placed on each plate of capacitor.

Solution:-

As we know charge on plate { +q and -q} , now we have find the charge according to given value

We know that Charge q = CV.

i.e

Charge q = (7.28μF)x(25.0V).

            q = 182μC.

Therefore the amount of charge on each plate will be q = 182μC.

i.e q₁ = + 182 μC and      q₂ = -182μC