Verified Solution • 2025–26 Session moderate

QPages 121-124: Q61: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q61 Solution: Two charged particles, with equal charges of , are brought from …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often multiply the $kq^2/r$ result by 2, thinking each charge contributes separately; remember the formula $U= k,q_1q_2/r$ already accounts for the single interaction between the two charges, so no extra factor of 2 is needed.

Two charged particles, with equal charges of 2.0×10−5 C2.0 \times 10^{-5} \text{ C}, are brought from infinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.

Step-by-Step Verified Solution

**Answer:** Increase in potential energy = 36 J. **STEP-BY-STEP Explanation:** q=2.0×10−5q = 2.0 \times 10^{-5} C each, final separation r=0.1r = 0.1 m.
U=kq2r=9×109×(2×10−5)20.1=36 JU = \frac{k q^2}{r} = \frac{9 \times 10^9 \times (2 \times 10^{-5})^2}{0.1} = 36 \text{ J}

All Questions in this Chapter

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