A parallel plate capacitor has plates of area 200 cm2 and separation 0.05 cm. It has been charged to a potential difference of 300 V, the energy of the capacitor is :

Question

A parallel plate capacitor has plates of area 200 cm2 and separation 0.05 cm. It has been charged to a potential difference of 300 V, the energy of the capacitor is : (A)  2.0× 10-6J (B)  2.4× 10-5 J (C)  0.8× 10-5J (D)  1.6× 10-5J

Tardigrade Admin · Accepted Answer

The capacitance of a parallel plate capacitor of plate area

A

and plate separated by distance d is given by

C = \frac{ε _{0} A}{d}

Given,

A = 200 c m^{2} = 200 \times 1 0^{- 4} m^{2}

,

d = 0.05 c m = 0.05 \times 1 0^{- 2} m

,

ε_{0} = 8.86 \times 1 0^{- 12} C^{2} / N - m^{2}

∴ C = \frac{8.86 \times 1 0 ^{- 12} \times 200 \times 1 0 ^{- 4}}{0.05 \times 1 0 ^{- 2}}

= 3.54 \times 1 0^{- 10} μ F

Energy stored in capacitor is

U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 3.54 \times 1 0^{- 10} \times 9 \times 1 0^{4}

= 1.6 \times 1 0^{- 5} J

Note: This energy resides in the electric field created between the plates of the charged capacitor.

Q. A parallel plate capacitor has plates of area $200 c m^{2}$ and separation $0.05 c m$ . It has been charged to a potential difference of $300 V$ , the energy of the capacitor is :

$2.0 \times 1 0^{- 6} J$

$2.4 \times 1 0^{- 5} J$

$0.8 \times 1 0^{- 5} J$

$1.6 \times 1 0^{- 5} J$

Q. A parallel plate capacitor has plates of area 200cm2 and separation 0.05cm. It has been charged to a potential difference of 300V, the energy of the capacitor is :

2.0×10−6J

2.4×10−5J

0.8×10−5J

1.6×10−5J

Q. A parallel plate capacitor has plates of area $200 c m^{2}$ and separation $0.05 c m$ . It has been charged to a potential difference of $300 V$ , the energy of the capacitor is :

$2.0 \times 1 0^{- 6} J$

$2.4 \times 1 0^{- 5} J$

$0.8 \times 1 0^{- 5} J$

$1.6 \times 1 0^{- 5} J$