Verified Solution • 2025–26 Session moderate

QPages 121-124: Q32.2: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q32.2 Solution: Two particles A and B, each carrying a charge Q, are held fixed …

⚠️ Common Exam Traps & Conceptual Pitfalls

When applying the binomial approximation for $x \ll d$, students frequently forget that the distance term is raised to the power of $3/2$ in Coulomb's law, leading to an incorrect linear term in the denominator. You must expand $(d^2/4 + x^2)^{3/2}$ correctly to show that the $x^2$ term vanishes, leaving a constant denominator proportional to $d^3$.

Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C of mass m and charge q is kept at the middle point of the line AB. Assuming x≪dx \ll d, show that this force is proportional to x.

Step-by-Step Verified Solution

**Answer:** F∝xF \propto x. **Step-BY-STEP Explanation:** For x≪dx \ll d, (d2/4+x2)3/2≈(d2/4)3/2=d3/8(d^2/4 + x^2)^{3/2} \approx (d^2/4)^{3/2} = d^3/8.
F≈2kQqxd3/8=16kQqd3x∝xF \approx \frac{2kQqx}{d^3/8} = \frac{16kQq}{d^3}x \propto x

All Questions in this Chapter

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