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QPages 121-124: Q30: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q30 Solution: A particle A with a charge of and a mass of 100 g is placed at…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently write the electric force as $F=kq^2/r$ and solve $r= kq^2/F$, overlooking the $1/r^2$ dependence; always remember $F=kq^2/r^2$, so the equilibrium distance is $r=\sqrt{kq^2/F}$, not a linear ratio.

A particle A with a charge of 2.0×10−6 C2.0 \times 10^{-6} \text{ C} and a mass of 100 g is placed at the bottom of a smooth inclined plane of inclination 30�. Where should another particle B, with the same charge and mass, be placed on the incline so that it may remain in equilibrium?

Step-by-Step Verified Solution

**Answer:** B should be placed at distance r=kq2mgsin⁡30�r = \sqrt{\frac{k q^2}{mg\sin 30�}} along the incline. **Step-BY-STEP Explanation:** For equilibrium on the incline, the electric force must balance the component of weight along the incline:
Fe=mgsin⁡30�=0.1×10×0.5=0.5 NF_e = mg\sin 30� = 0.1 \times 10 \times 0.5 = 0.5 \text{ N}
r=kq2Fe=9×109×(2×10−6)20.5=0.27 mr = \sqrt{\frac{k q^2}{F_e}} = \sqrt{\frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{0.5}} = 0.27 \text{ m}

All Questions in this Chapter

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