A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

E₁=1/2 x CV²

= 1/2   x (600 x 10⁻¹²) x (200)² J = 1.2 x 10⁻⁵ J

If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C_eq) of the combination is given by,

1/C_eq=1/C ⁺ 1/C

=> 1/C_eq =1/600 + 1/600=2/600 =1/300

=> C_eq =300 pF

New electrostatic energy can be calculated as

E₂=1/2 C_eq V²

=1/2 x 300 x (200)² J = 0.6 x 10⁻⁵ J

Loss in electrostatic energy = E₁ – E2

= 1.2 x 10⁻⁵ – 0.6 x 10⁻⁵ J = 0.6 x 10⁻⁵J = 6x 10⁻⁶J

Therefore, the electrostatic energy lost in the process is 6 x 10^-6 J .