Question:medium
Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is: ____.
Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is: ____.
Show Hint
When two \textbf{identical} capacitors are connected (one charged, one uncharged), exactly half of the initial energy is always lost. Initial energy was $\frac{1}{2}CV^2$, so loss is $\frac{1}{4}CV^2$.
Updated On: May 28, 2026
- 1.0 J
- 0.5 J
- 1.0 $\times$ 10⁻⁶ J
- 0.5 $\times$ 10⁻⁶ J
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Topic:
This problem falls under "Electrostatic Potential and Capacitance." It describes the common scenario of charge sharing between capacitors. When a charged capacitor is connected to another capacitor, charge flows until both reach a common potential. This movement of charge through wires (which have some resistance) results in energy being lost as heat.
Step 2: Key Formulas and Approach:
The general formula for energy loss ($\Delta U$) when two capacitors are connected is: \[ \Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)} \] Alternatively, you can calculate the initial energy, the common potential, the final energy, and then find the difference.
Step 3: Detailed Explanation:
Identify given values: $C_1 = C_2 = 200 \text{ pF} = 200 \times 10^{-12} \text{ F}$. Initial voltages: $V_1 = 100 \text{ V}$ and $V_2 = 0 \text{ V}$ (uncharged).
Simplify for equal capacitances: If $C_1 = C_2 = C$, the loss formula becomes: \[ \Delta U = \frac{C^2 (V_1 - 0)^2}{2(2C)} = \frac{C \cdot V_1^2}{4} \]
Perform the calculation: \[ \Delta U = \frac{(200 \times 10^{-12} \text{ F}) \times (100 \text{ V})^2}{4} \] \[ \Delta U = \frac{200 \times 10^{-12} \times 10^4}{4} \] \[ \Delta U = \frac{200 \times 10^{-8}}{4} = 50 \times 10^{-8} \text{ J} \]
Convert to scientific notation: \[ 50 \times 10^{-8} \text{ J} = 0.5 \times 10^{-6} \text{ J} \]
This loss accounts for exactly 50% of the initial stored energy, which was $\frac{1}{2} C_1 V_1^2 = 1.0 \times 10^{-6} \text{ J}$.
Step 4: Final Answer:
The amount of electrostatic energy lost is 0.5 $\times$ 10⁻⁶ J.
This problem falls under "Electrostatic Potential and Capacitance." It describes the common scenario of charge sharing between capacitors. When a charged capacitor is connected to another capacitor, charge flows until both reach a common potential. This movement of charge through wires (which have some resistance) results in energy being lost as heat.
Step 2: Key Formulas and Approach:
The general formula for energy loss ($\Delta U$) when two capacitors are connected is: \[ \Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)} \] Alternatively, you can calculate the initial energy, the common potential, the final energy, and then find the difference.
Step 3: Detailed Explanation:
Identify given values: $C_1 = C_2 = 200 \text{ pF} = 200 \times 10^{-12} \text{ F}$. Initial voltages: $V_1 = 100 \text{ V}$ and $V_2 = 0 \text{ V}$ (uncharged).
Simplify for equal capacitances: If $C_1 = C_2 = C$, the loss formula becomes: \[ \Delta U = \frac{C^2 (V_1 - 0)^2}{2(2C)} = \frac{C \cdot V_1^2}{4} \]
Perform the calculation: \[ \Delta U = \frac{(200 \times 10^{-12} \text{ F}) \times (100 \text{ V})^2}{4} \] \[ \Delta U = \frac{200 \times 10^{-12} \times 10^4}{4} \] \[ \Delta U = \frac{200 \times 10^{-8}}{4} = 50 \times 10^{-8} \text{ J} \]
Convert to scientific notation: \[ 50 \times 10^{-8} \text{ J} = 0.5 \times 10^{-6} \text{ J} \]
This loss accounts for exactly 50% of the initial stored energy, which was $\frac{1}{2} C_1 V_1^2 = 1.0 \times 10^{-6} \text{ J}$.
Step 4: Final Answer:
The amount of electrostatic energy lost is 0.5 $\times$ 10⁻⁶ J.
Was this answer helpful?
3
