700 pF capacitor is charged by 50 V battery. Electrostatic energy is stored by it will be :

Question

700 pF capacitor is charged by 50 V battery. Electrostatic energy is stored by it will be : (A)  17.0× 10-8J (B)  13.0× 10-9J  (C)  8.7× 10-7J  (D)  6.7× 10-7J

Tardigrade Admin · Accepted Answer

Here: Capacitance C=700 pF =700 × 10-12 F Source voltage V=50 V Electrostatic energy is given by E =(1/2) C V2 =0.5 × 700 × 10-12 ×(50)2 =8.7 × 10-7 J

Q. $700 pF$ capacitor is charged by $50 V$ battery. Electrostatic energy is stored by it will be :

$17.0 \times 1 0^{- 8} J$

$13.0 \times 1 0^{- 9} J$

$8.7 \times 1 0^{- 7} J$

$6.7 \times 1 0^{- 7} J$

Here : Capacitance
$C = 700 pF = 700 \times 1 0^{- 12} F$
Source voltage $V = 50 V$
Electrostatic energy is given by
$E = \frac{1}{2} C V^{2}$
$= 0.5 \times 700 \times 1 0^{- 12} \times (50)^{2}$
$= 8.7 \times 1 0^{- 7} J$

Q. 700pF capacitor is charged by 50V battery. Electrostatic energy is stored by it will be :

17.0×10−8J

13.0×10−9J

8.7×10−7J

6.7×10−7J