Question

a capacitor carries a charge of 0.1 C at 5 volt its capacitance is​

Answer 1

Given :

Potential difference (V) = 5 V

Charge (Q) = 0.1 C

To Find :

Capacitance of the capacitor

Solution :

The capacitance or capacity of a conductor is defined as the charge required to raise its potential by unity.

         Capacitance (C) = $\frac{Charge (Q)}{Potential Difference (V)}$

∴ Capacitance of the capacitor (C) = $\frac{0.1}{5}$

                                                         = 0.02 F

Hence, the capacitance of the capacitor carrying a charge of 0.1 C and having a potential difference of 5 V is found to be 0.02 F.

Answer 2

Answer:

The capacitance is 0.02 F

Explanation:

Here we have been given to find the capacitance of the given capacitance which is carrying a charge of 0.1 C and is being operated at a 5 V voltage source.

Now as we know that the capacitance is designated as the amount of charge that can accumulate inside the given capacitor at the given potential across its plates.

The formula for calculating the capacitance (C) is as below:

C = $\frac{Q}{V}$

Here C is the capacitance, Q is the charge and V is the given potential difference.

We have,

V = 5 volts

Q = 0.1 C

⇒ $C = \frac{0.1 }{5}$ F

⇒ C = 0.02 F

Therefore the capacitance is found to be 0.02 F