Question:
The work done in placing a charge of \( 8 \times 10^{-18} \) coulomb on a capacitor of capacitance 100 microfarad is:
The work done in placing a charge of \( 8 \times 10^{-18} \) coulomb on a capacitor of capacitance 100 microfarad is:
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The formula \( W = \frac{Q^2}{2C} \) is useful for calculating the energy stored in a capacitor when charge and capacitance are known.
Updated On: Feb 4, 2025
- \( 3.1 \times 10^{-26} \) joule
- \( 4 \times 10^{-10} \) joule
- \( 32 \times 10^{-32} \) joule
- \( 16 \times 10^{-32} \) joule
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The Correct Option is C
Solution and Explanation
Step 1: Formula for work done in charging a capacitor
The work done \( W \) in charging a capacitor is given by: \[ W = \frac{Q^2}{2C} \] where: \( Q = 8 \times 10^{-18} \) C (charge), \( C = 100 \times 10^{-6} \) F (capacitance).
Step 2: Substituting values
\[ W = \frac{(8 \times 10^{-18})^2}{2 \times (100 \times 10^{-6})} \] \[ = \frac{64 \times 10^{-36}}{2 \times 100 \times 10^{-6}} \] \[ = \frac{64 \times 10^{-36}}{200 \times 10^{-6}} \] \[ = 32 \times 10^{-32} { joule} \]
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