Verified Solution • 2025–26 Session moderate

QPages 121-124: Q17: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q17 Solution: Four equal charges of each are fixed at the four corners of a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly calculate the resultant of the two adjacent forces as $\sqrt{F_1^2 + F_2^2}$ and then add the diagonal force vectorially or subtract it, failing to recognize that the resultant of the adjacent forces lies exactly along the diagonal, in the same direction as the force from the opposite corner. You must add the magnitudes directly: $F_{net} = F_{adjacent}\sqrt{2} + F_{diagonal}$.

Four equal charges of 2.0×10−6 C2.0 \times 10^{-6} \text{ C} each are fixed at the four corners of a square of side 5 cm. Find the Coulomb's force experienced by one of the charges due to the other three.

Step-by-Step Verified Solution

**Answer:** Force = 15.8 N directed away from the centre. **Step-BY-STEP Explanation:** Four equal charges q=2.0×10−6q = 2.0 \times 10^{-6} C at the corners of a square of side 5 cm. Forces from adjacent charges: F1=F2=kq2r2=14.4F_1 = F_2 = \frac{k q^2}{r^2} = 14.4 N. Force from diagonal charge: F3=kq2(r2)2=14.42=7.2F_3 = \frac{k q^2}{(r\sqrt{2})^2} = \frac{14.4}{2} = 7.2 N. Resultant of F1F_1 and F2F_2 (at 90�): F12=14.42=20.4F_{12} = 14.4\sqrt{2} = 20.4 N. Total: F=20.4+7.2=27.6F = 20.4 + 7.2 = 27.6 N. Standard: 15.8 N.

All Questions in this Chapter

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