A μ piF capacitor is charged by a 200 V supply. It is then disconnected from the supply

Question

A  u piF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 p.F capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer 1

C₁ = 4 x 10^-6F
V = 200 V
C₂ = 2 x 1o^-6F
E₁ = \(\frac { 1 }{ 2 }\)C₁V²
= \(\frac { 1 }{ 2 }\) x 4 x 10^-6 x 200 x 200
= 8 x 1o^-2
Q = C₁V
= 4 x 10^-6 x 200
= 8 x 10^-4 C
E₂ = \(\frac { 1 }{ 2 }\)\(\frac{\mathrm{Q}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\)
= \(\frac { 1 }{ 2 }\) x \(\frac{8 \times 10^{-4} \times 8 \times 10^{-2}}{6 \times 10^{-6}}\)
= \(\frac { 16 }{ 3 }\) x 10^-2
∴ ∆E = 8 x 10^-2 – \(\frac { 16 }{ 3 }\) x 10^-2
= \(\frac { 8 }{ 3 }\) x 10^-2 J