Verified Solution • 2025–26 Session moderate

QPages 121-124: Q25: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q25 Solution: Two identical pith balls, each carrying a charge q, are…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently mistake the separation distance between the charges as $l$ or $l\sin\theta$ instead of the correct chord length $2l\sin\theta$. Always draw the geometry to identify that the horizontal distance from the vertical axis to one ball is $l\sin\theta$, making the total separation $2l\sin\theta$.

Two identical pith balls, each carrying a charge q, are suspended from a common point by two strings of equal length l. Find the mass of each ball if the angle between the strings is

2\theta

in equilibrium.

Step-by-Step Verified Solution

**Answer:**

m = \frac{k q^2}{4g l^2 \sin^3\theta}

. **Step-BY-STEP Explanation:** For equilibrium:

T\sin\theta = F_e

,

T\cos\theta = mg

.

\tan\theta = \frac{F_e}{mg} = \frac{k q^2}{m g (2l\sin\theta)^2}

. Solving:

m = \frac{k q^2}{4g l^2 \sin^3\theta \cos\theta}

.

All Questions in this Chapter

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