Question:
When a capacitor of $9\,pF$ is connected to a battery, the electrostatic energy stored in the capacitor is $18 \times 10^{-8}J$. The quantity of charge stored in the capacitor is:
When a capacitor of $9\,pF$ is connected to a battery, the electrostatic energy stored in the capacitor is $18 \times 10^{-8}J$. The quantity of charge stored in the capacitor is:
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Use $U = \frac{Q^2}{2C}$ when voltage is not given.
Updated On: Apr 24, 2026
- $1.2\,nC$
- $1.8\,nC$
- $2.7\,nC$
- $3.6\,nC$
- $2.4\,nC$
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The Correct Option is B
Solution and Explanation
Concept:
• Energy stored: \[ U = \frac{Q^2}{2C} \]
Step 1: Given
\[ C = 9\times10^{-12}, \quad U = 18\times10^{-8} \]
Step 2: Find charge
\[ Q = \sqrt{2CU} \] \[ = \sqrt{2 \cdot 9\times10^{-12} \cdot 18\times10^{-8}} \] \[ = \sqrt{324\times10^{-20}} = 18\times10^{-10} \] \[ = 1.8\times10^{-9}C = 1.8\,nC \] Final Conclusion:
Option (B)
• Energy stored: \[ U = \frac{Q^2}{2C} \]
Step 1: Given
\[ C = 9\times10^{-12}, \quad U = 18\times10^{-8} \]
Step 2: Find charge
\[ Q = \sqrt{2CU} \] \[ = \sqrt{2 \cdot 9\times10^{-12} \cdot 18\times10^{-8}} \] \[ = \sqrt{324\times10^{-20}} = 18\times10^{-10} \] \[ = 1.8\times10^{-9}C = 1.8\,nC \] Final Conclusion:
Option (B)
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