Question:
Two point charges $+10\mu C$ and $4\mu C$ are placed 10 cm apart in air. The work required to be done to bring them 2 cm closer is ($\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ SI units)
Two point charges $+10\mu C$ and $4\mu C$ are placed 10 cm apart in air. The work required to be done to bring them 2 cm closer is ($\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ SI units)
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Work is positive when bringing like charges closer, as you are moving against the electrostatic repulsion.
Updated On: Apr 30, 2026
- 0.65 J
- 0.9 J
- 1.2 J
- 2.3 J
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The Correct Option is B
Solution and Explanation
Step 1: Formula
Work Done $W = \Delta U = k q_1 q_2 (\frac{1}{r_2} - \frac{1}{r_1})$
Step 2: Given Values
$r_1 = 10 \text{ cm} = 0.1 \text{ m}$
$r_2 = 10 - 2 = 8 \text{ cm} = 0.08 \text{ m}$
Step 3: Calculation
$W = 9 \times 10^9 \times 10 \times 10^{-6} \times 4 \times 10^{-6} \times (\frac{1}{0.08} - \frac{1}{0.1})$
$W = 0.36 \times (12.5 - 10) = 0.36 \times 2.5 = 0.9 \text{ J}$
Step 4: Conclusion
The work required is 0.9 J.
Final Answer:(B)
Work Done $W = \Delta U = k q_1 q_2 (\frac{1}{r_2} - \frac{1}{r_1})$
Step 2: Given Values
$r_1 = 10 \text{ cm} = 0.1 \text{ m}$
$r_2 = 10 - 2 = 8 \text{ cm} = 0.08 \text{ m}$
Step 3: Calculation
$W = 9 \times 10^9 \times 10 \times 10^{-6} \times 4 \times 10^{-6} \times (\frac{1}{0.08} - \frac{1}{0.1})$
$W = 0.36 \times (12.5 - 10) = 0.36 \times 2.5 = 0.9 \text{ J}$
Step 4: Conclusion
The work required is 0.9 J.
Final Answer:(B)
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