Question

A 6 μF capacitor is charged by a 100V supply. It is then disconnected from the supply and μ is connected to another uncharged 3μF capacitor. How much electrostatic energy of the first capacitor is dissipated in the form of heat and electromagnetic radiation?

(a) 3 x 10^-2 J

(b) 2 x 10^-2 J

(c) 1 x 10^-2 J

(d) 4 x 10^-2 J

I got this question in exam.

Origin of the question is Energy Stored in a Capacitor topic in division Electrostatic Potential and Capacitance of Physics – Class 12

Answer 1

Right choice is (c) 1 x 10^-2 J

The best explanation: Given: C1 = 6μF; V1 = 100 V

Initial energy stored (U1) = \(\frac {1}{2}\) C1V1^2 = \(\frac {1}{2}\) × 6 × 10^-6 × 100 × 100

                                             = 3 × 10^-2 J

Potential (V) = \(\frac {C_1V_1}{(C_1+C_2)}\)  

                       = 6 × 10^-6 × \(\frac {100}{(6+3)}\) × 10^-6

                       = \(\frac {600}{9}\) V

Final energy (U2) = \(\frac {1}{2}\) (C1 + C2) × V^2

                               = \(\frac {1}{2}\) × 9 × 10^-6 × \(\frac {1}{2} \times \frac {600}{9} \times \frac {600}{9}\)

                               = 2 × 10^-2 J

Therefore, the energy dissipated in the form of heat and electromagnetic radiation is:

U2 – U1 = 3 × 10^-2 – 2 × 10^-2

U2 – U1 = 1 × 10^-2 J