Question

A capacitor of capacitance $8\mu\text{F}$ is fully charged by connecting it to a source of $200\text{ V}$. It is then disconnected from the supply and connected to an uncharged capacitor of capacitance $4\mu\text{F}$. The electrostatic energy lost in this sharing process is:

• A. $5.33 \times 10^{-2}\text{ J}$
• B. $21.34 \times 10^{-2}\text{ J}$
• C. $10.67 \times 10^{-2}\text{ J}$
• D. $3.53 \times 10^{-3}\text{ J}$

Hint:

Using the formula $\Delta U = \frac{1}{2}\frac{C_1C_2}{C_1+C_2}V^2$ allows you to solve the problem in a single line, avoiding the multi-step process of finding the total initial energy, calculating the common potential $V_c = \frac{C_1V_1}{C_1+C_2}$, and computing the final system energy.

Answer 1

The Correct Option is A

Understanding the Concept: When a charged capacitor $C_1$ at potential $V_1$ is connected in parallel with an uncharged capacitor $C_2$ ($V_2 = 0$), charges redistribute until they reach a common potential. This movement of charge through wire resistance dissipates potential energy as heat. The net energy loss ($\Delta U$) can be evaluated directly using: \[ \Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2 \]
Step 1: Substitute values into the energy loss equation.
We are given:
$C_1 = 8\mu\text{F} = 8 \times 10^{-6}\text{ F}$
$C_2 = 4\mu\text{F} = 4 \times 10^{-6}\text{ F}$
$V_1 = 200\text{ V}$, $V_2 = 0\text{ V}$
Plugging these variables into the expression: \[ \Delta U = \frac{1}{2} \cdot \frac{(8 \times 10^{-6}) \cdot (4 \times 10^{-6})}{(8 + 4) \times 10^{-6}} \cdot (200 - 0)^2 \] \[ \Delta U = \frac{1}{2} \cdot \frac{32 \times 10^{-12}}{12 \times 10^{-6}} \cdot 40000 \]
Step 2: Simplify the numbers.
\[ \Delta U = \frac{1}{2} \cdot \frac{8}{3} \times 10^{-6} \cdot 40000 = \frac{4}{3} \times 10^{-6} \cdot 40000 \] \[ \Delta U = \frac{160000}{3} \times 10^{-6} = 53333.33 \times 10^{-6}\text{ J} = 5.33 \times 10^{-2}\text{ J} \]