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QPages 121-124: Q6: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q6 Solution: Two charges and are placed at a separation of 10 cm. Where…

Two charges 2.0×10−6 C2.0 \times 10^{-6} \text{ C} and 1.0×10−6 C1.0 \times 10^{-6} \text{ C} are placed at a separation of 10 cm. Where should a third charge be placed, such that it experiences no net force due to these charges?

Step-by-Step Verified Solution

**Answer:** The third charge should be placed 5.9 cm from the smaller charge (1.0 _ 10__ C) or 4.1 cm from the larger charge (2.0 _ 10__ C). **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. Let the third charge be at distance xx from q1q_1 (between the charges). For net force zero:
kq1qx2=kq2q(0.1−x)2\frac{k q_1 q}{x^2} = \frac{k q_2 q}{(0.1 - x)^2}
2x2=1(0.1−x)2\frac{2}{x^2} = \frac{1}{(0.1 - x)^2}
2(0.1−x)=x⇒x=2×0.11+2=0.0586 m=5.9 cm\sqrt{2}(0.1 - x) = x \Rightarrow x = \frac{\sqrt{2} \times 0.1}{1 + \sqrt{2}} = 0.0586 \text{ m} = 5.9 \text{ cm}
Distance from smaller charge = 5.9 cm, from larger charge = 4.1 cm .

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