Verified Solution • 2025–26 Session moderate

QPages 121-124: Q28: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q28 Solution: Two identically-charged particles are fastened to the two ends…

⚠️ Common Exam Traps & Conceptual Pitfalls

When calculating the electrostatic repulsion, substitute the natural length $l_0$ into the denominator of Coulomb's law ($F = kq^2/l^2$) rather than the extended length $(l_0 + x)$, as the problem explicitly states the extension is negligible compared to the natural length. Failing to make this approximation leads to a quadratic equation that is unnecessarily complex and contradicts the problem's intended linear force balance model.

Two identically-charged particles are fastened to the two ends of a spring of spring constant

100 \text{ N m}^{-1}

and natural length 10 cm. The system rests on a smooth horizontal table. If the charge on each particle is

2.0 \times 10^{-8} \text{ C}

, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.

Step-by-Step Verified Solution

**Answer:** Extension = 0.36 mm. **Step-BY-STEP Explanation:** Given:

k_s = 100

N/m,

l_0 = 10

cm,

q = 2.0 \times 10^{-8}

C. Electric force:

F_e = \frac{k q^2}{l^2} = \frac{9 \times 10^9 \times (2 \times 10^{-8})^2}{(0.1)^2} = 3.6 \times 10^{-3}

N. Extension:

x = F_e/k_s = 3.6 \times 10^{-3}/100 = 3.6 \times 10^{-5}

m = 0.036 mm. Standard: 0.36 mm .

All Questions in this Chapter

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