Question

A 50 μF capacitor has a potential difference of 8 V across it. The charge on capacitor is​

Answer 1

Step-by-step explanation:

Q=CV

Q=50×10^-6 • 8

Q=400×10^-6

Answer 2

Given,

A 50 μF capacitor has a potential difference of 8 V across it.

To Find,

The charge on the capacitor is​ =?

Solution,

We can solve the question as follows:

It is given to us that a 50 μF capacitor has a potential difference of 8 V across it. We have to find the charge on the capacitor.

The charge, Q on a capacitor is equal to the capacitance of the capacitor multiplied by the potential difference across it.

$Q = CV$

Where, $Q = Charge\: stored$

            $C = Capacitance$

           $V = Potential\: difference$

Substituting C = 50 μF and V = 8 V in the above formula,

Since the capacitance is not in S.I. units, unit conversion is required.

C = 50 μF = 50 x 10^-6 F       (1 μF = 10^-6 F)

Now,

$Q = (50*10^{-6} )*8$

$Q = 400 * 10^{-6} = 4 * 10^{-4} C$

Hence, the charge on the capacitor is​ 4 x 10^-4 C.