A capacitor of capacitance 700 pF is charged by 100 V battery. The electrostatic energy stored by the capacitor is

Question

A capacitor of capacitance 700 pF is charged by 100 V battery. The electrostatic energy stored by the capacitor is (A) 2.5 × 10-8 J (B) 3.5 × 10-6 J (C) 2.5 × 10-4 J (D) 3.5 × 10-4 J

Tardigrade Admin · Accepted Answer

Here, C=700 PF =700×10-12 F, V=100 V Energy stored U=(1/2)CV2=(1/2)×700×10-12×(100)2 =3.5×10-6 J

Q. A capacitor of capacitance $700 pF$ is charged by $100 V$ battery. The electrostatic energy stored by the capacitor is

$2.5 \times 1 0^{- 8} J$

$3.5 \times 1 0^{- 6} J$

$2.5 \times 1 0^{- 4} J$

$3.5 \times 1 0^{- 4} J$

Here, $C = 700 PF = 700 \times 1 0^{- 12} F$ ,
$V = 100 V$
Energy stored
$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 700 \times 1 0^{- 12} \times (100)^{2}$
$= 3.5 \times 1 0^{- 6} J$

Q. A capacitor of capacitance 700pF is charged by 100V battery. The electrostatic energy stored by the capacitor is

2.5×10−8J

3.5×10−6J

2.5×10−4J

3.5×10−4J