Verified Solution • 2025–26 Session moderate

QPages 121-124: Q7: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q7 Solution: Suppose the second charge in the previous problem is . Locate…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume the equilibrium point lies between opposite charges and write $kq_1q/(x)^2 = -kq_2q/(r-x)^2$, mixing signs and getting no solution. Remember that for opposite signs the forces point opposite ways, so the zero‑force point is *outside* the pair, on the side of the smaller‑magnitude charge; set the magnitudes equal and use distances measured from that side (e.g., $k|q_1|/(x+r)^2 = k|q_2|/x^2$).

Suppose the second charge in the previous problem is −1.0×10−6 C-1.0 \times 10^{-6} \text{ C}. Locate the position where a third charge will not experience a net force.

Step-by-Step Verified Solution

**Answer:** The third charge should be placed 34.1 cm from the larger charge on the side of the smaller charge. **Step-BY-Step Explanation:** Given: q1=2.0×10−6q_1 = 2.0 \times 10^{-6} C, q2=−1.0×10−6q_2 = -1.0 \times 10^{-6} C, r=10r = 10 cm. Since the charges are opposite, the zero-force point lies outside the charges, on the side of the smaller charge (the negative charge). Let it be at distance xx from the negative charge q2q_2.
kq1q(x+0.1)2=kq2qx2\frac{k q_1 q}{(x + 0.1)^2} = \frac{k q_2 q}{x^2}
2(x+0.1)2=1x2\frac{2}{(x + 0.1)^2} = \frac{1}{x^2}
2x=x+0.1⇒x=0.12−1=0.2414 m=24.1 cm\sqrt{2}x = x + 0.1 \Rightarrow x = \frac{0.1}{\sqrt{2} - 1} = 0.2414 \text{ m} = 24.1 \text{ cm}
Distance from larger charge = 24.1+10=34.124.1 + 10 = 34.1 cm .

All Questions in this Chapter

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