Question

A capacitor of capacitance 2micro farad is charged to a potential of 10V calculate the energy stored in it​

Answer 1

Answer: $10^{-4} J$

Explanation:

Capacitance C = 2 μF = 2 × $10^{-6}$ F

Potential difference, V = 10 V

The energy stored in the capacitor $E= \frac{1}{2} C V^{2}$ = $\frac{1}{2}$ x 2 x $10^{-6}$ x $(10)^{2}$

                                                                       = 0.0001 J

                                                                       = $10^{-4} J$

Answer 2

Answer:

A capacitor of capacitance 2micro farad is charged to a potential of 10V. The energy stored in it​ is $1\ J$

Explanation:

• Let us consider a capacitor of capacitance C and with a potential difference between the capacitor plates V.

• Electric potential energy (E) stored in the capacitor  can be expressed as,

            $E=\frac{1}{2} C V^{2}$    ...(1)

In the question, it is given that,

$V= 10\ V$

$C =2\ \mu F = 2\times10^{-6} F$

Substitute these  values into equation (1)

We get,

$E=\frac{1}{2}\times 2\times10^{-6} \times 10^{2} = 1 \ J$

Thus,

The energy stored between the plates of the capacitor  will be   $1\ J$