Question:
Two point charges \(q_1 = 6 μ\text{C}\) and \(q_2 = 4 μ\text{C}\) are kept at points A and B in air where distance \(AB = 10\) cm. What is the increase in potential energy of the system when \(q_2\) is moved towards \(q_1\), by 2 cm ? \((\frac{1}{4πε_0} = 9\times 10^9 \text{SI units})\)
Two point charges \(q_1 = 6 μ\text{C}\) and \(q_2 = 4 μ\text{C}\) are kept at points A and B in air where distance \(AB = 10\) cm. What is the increase in potential energy of the system when \(q_2\) is moved towards \(q_1\), by 2 cm ? \((\frac{1}{4πε_0} = 9\times 10^9 \text{SI units})\)
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Potential energy of two charges is k q1 q2 / r; find the change from 10 cm to 8 cm.
Updated On: Oct 1, 2026
- \(21.6\) J
- \(216\) J
- \(0.54\) J
- \(54\) J
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The potential energy of a pair of point charges is \(U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}\).
Step 2: Initial and final energies:
\(q_1q_2 = 6\times10^{-6}\times4\times10^{-6} = 24\times10^{-12}\) C\(^2\).
Initially \(r = 0.10\) m: \(U_1 = \dfrac{9\times10^9\times24\times10^{-12}}{0.10} = 2.16\) J.
After moving 2 cm closer \(r = 0.08\) m: \(U_2 = \dfrac{9\times10^9\times24\times10^{-12}}{0.08} = 2.70\) J.
Step 3: Change:
\[ \Delta U = 2.70 - 2.16 = 0.54\ \text{J} \]
Step 4: Check:
Option (C). Options (A), (B), (D) are off by powers of ten or use the wrong separation.
Final Answer:
U changes from 2.16 J to 2.70 J. \[ \boxed{\text{(C) }0.54\ \text{J}} \]
The potential energy of a pair of point charges is \(U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}\).
Step 2: Initial and final energies:
\(q_1q_2 = 6\times10^{-6}\times4\times10^{-6} = 24\times10^{-12}\) C\(^2\).
Initially \(r = 0.10\) m: \(U_1 = \dfrac{9\times10^9\times24\times10^{-12}}{0.10} = 2.16\) J.
After moving 2 cm closer \(r = 0.08\) m: \(U_2 = \dfrac{9\times10^9\times24\times10^{-12}}{0.08} = 2.70\) J.
Step 3: Change:
\[ \Delta U = 2.70 - 2.16 = 0.54\ \text{J} \]
Step 4: Check:
Option (C). Options (A), (B), (D) are off by powers of ten or use the wrong separation.
Final Answer:
U changes from 2.16 J to 2.70 J. \[ \boxed{\text{(C) }0.54\ \text{J}} \]
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