Verified Solution • 2025–26 Session moderate

QPages 121-124: Q20: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q20 Solution: Ten positively-charged particles are kept fixed on the x-axis…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug $x_n = n\times10\text{ cm}$ directly into Coulomb's law without converting to meters, forgetting that $(0.1n\,\text{m})^2 =0.01 n^2$; this omission introduces a factor of 100 error in the force. Always convert distances to SI units before squaring them.

Ten positively-charged particles are kept fixed on the x-axis at points x = 10 cm, 20 cm, 30 cm, ...., 100 cm. The first particle has a charge 1.0×10−8 C1.0 \times 10^{-8} \text{ C}, the second 8×10−8 C8 \times 10^{-8} \text{ C}, the third 27×10−8 C27 \times 10^{-8} \text{ C} and so on. The tenth particle has a charge 1000×10−8 C1000 \times 10^{-8} \text{ C}. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Step-by-Step Verified Solution

**Answer:** Force = 4.5 _ 10_ N. **Step-BY-STEP Explanation:** The charges are qn=n3×10−8q_n = n^3 \times 10^{-8} C at xn=n×10x_n = n \times 10 cm = n×0.1n \times 0.1 m for n=1,2,...,10n = 1, 2, ..., 10. Force on 1 C charge at origin:
F=∑n=110kqn×1xn2=∑n=1109×109×n3×10−8(0.1n)2=9×103∑n=110n=9×103×55=4.95×105 NF = \sum_{n=1}^{10} \frac{k q_n \times 1}{x_n^2} = \sum_{n=1}^{10} \frac{9 \times 10^9 \times n^3 \times 10^{-8}}{(0.1n)^2} = 9 \times 10^3 \sum_{n=1}^{10} n = 9 \times 10^3 \times 55 = 4.95 \times 10^5 \text{ N}
Standard: 4.5 _ 10_ N.

All Questions in this Chapter

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