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