Question

A capacitor of capacitance 20micro farad is charged to a potential of 500v. Calculate the charge and energy stored in a capacitor

Answer 1

Answer:

We know charge=capacity× voltage.

So final voltage =

C

net

Q

net

Q

net

=Q

1

+Q

2

=C

1

V

1

+C

2

V

2

In parallel the C

net

=C

1

+C

2

=30μF

So putting Corresponding Values we get final voltage as 400V

Option B is correct.

Answer 2

Answer:

A capacitor of capacitance 20micro farad is charged to a potential of 500v.

• The energy stored between the plates of the capacitor  will be   $2.5\ J$

• The charge stored = $0.01\ C$

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)

• The charge (Q) of the capacitor depends upon the capacitance and the applied voltage given by the formula

            $Q = CV$       ...(2)

In the question, it is given that,

$V= 500\ V$

$C =20\ \mu F = 20\times10^{-6} F$

Substitute these  values into equation (1)

We get,

$E=\frac{1}{2}\times 20\times10^{-6} \times 500^{2} = 2.5 \ J$

Substitute values of C and V to equation (2) to obtain the value of Q.

$Q = 20\times 10^-6\times 500 = 0.01\ C$

Thus,

The energy stored between the plates of the capacitor  will be   $2.5\ J$

The charge stored = $0.01\ C$