Given :
Capacitance (C) = 3 nF = 3×10⁻⁹ F
Potential Difference = 400 V
To Find :
Energy stored in the condenser
Solution :
Energy stored in a charged condenser can be given as -
Energy (E) = [tex]\frac{1}{2} CV^2[/tex]
where, C = Capacitance; V = Potential difference
∴ Energy stored in the condenser (E) = [tex]\frac{1}{2}[/tex] × 3 ×10⁻⁹ ×400
= 6 × 10⁻⁷ J
Therefore, the energy stored in the condenser of capacitance 3 nF is found to be 6 × 10⁻⁷ J.