Question

A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is _____ μ.J

Answer 1

Correct Answer: 6

To determine the electrostatic energy lost when a charged capacitor is connected to another uncharged capacitor, we start by calculating the initial and final energies.

Step 1: Calculate Initial Electrostatic Energy

The initial energy stored in a capacitor is given by: \( E_{initial} = \frac{1}{2}CV^2 \). For a 600 pF capacitor (converted to farads as \( 600 \times 10^{-12} \) F) charged to 200 V:

\( E_{initial} = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = \frac{1}{2} \times 600 \times 10^{-12} \times 40000 \)

\( E_{initial} = 12 \times 10^{-6} \) J = 12 μJ

Step 2: Calculate Final Electrostatic Energy

When the charged capacitor is connected to an uncharged identical capacitor, they share the total charge. The final voltage across each capacitor is half the initial voltage:

Final voltage \( V_f = \frac{200}{2} = 100 \) V

Total capacitance for the two capacitors in parallel is \( C_{total} = 600 \times 10^{-12} + 600 \times 10^{-12} = 1200 \times 10^{-12} \) F

Final energy stored \( E_{final} = \frac{1}{2}C_{total}V_f^2 \)

\( E_{final} = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times 10000 \)

\( E_{final} = 6 \times 10^{-6} \) J = 6 μJ

Step 3: Calculate Energy Lost

Energy lost = Initial energy - Final energy:

Energy lost = 12 μJ - 6 μJ = 6 μJ

Conclusion: The electrostatic energy lost in the process is 6 μJ, which is within the given range of 6,6.