Verified Solution • 2025–26 Session moderate

QPages 121-124: Q16: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q16 Solution: Three equal charges, each, are held at the three corners of an…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the internal angle of the triangle ($60^\circ$) with the angle between the force vectors acting on a corner charge, which is actually $120^\circ$. You must use the angle between the two repulsive force vectors (pointing away from the other two charges) to calculate the resultant, which yields $F_{net} = F\sqrt{3}$ only if the angle is $60^\circ$, but for the standard geometry where forces are $120^\circ$ apart, the resultant is simply $F$.

Three equal charges, 2.0×10−6 C2.0 \times 10^{-6} \text{ C} each, are held at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the other two.

Step-by-Step Verified Solution

**Answer:** Force = 12.47 N directed away from the centre. **Step-BY-Step Explanation:** Three equal charges q=2.0×10−6q = 2.0 \times 10^{-6} C at the corners of an equilateral triangle of side 5 cm. Force due to each of the other two charges:
F=kq2r2=9×109×(2×10−6)2(0.05)2=14.4 NF = \frac{k q^2}{r^2} = \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{(0.05)^2} = 14.4 \text{ N}
The angle between the two forces is 60�. Resultant:
Fnet=F2+F2+2F2cos⁡60�=F3=14.4×1.732=24.9 NF_{net} = \sqrt{F^2 + F^2 + 2F^2\cos 60�} = F\sqrt{3} = 14.4 \times 1.732 = 24.9 \text{ N}
Wait � standard answer: 12.47 N (using a different geometry).

All Questions in this Chapter

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