Verified Solution • 2025–26 Session moderate

QPages 121-124: Q26: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q26 Solution: A particle with a charge of is placed directly below and at a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently forget to convert the separation distance from centimeters to meters before substituting it into Coulomb's law, leading to an error of $10^4$ in the final charge value.

A particle with a charge of 2.0×10−4 C2.0 \times 10^{-4} \text{ C} is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. The mass of the bob is 100 g. What charge should the bob be given so that the string becomes loose?

Step-by-Step Verified Solution

**Answer:** Charge on bob = 5.4 _ 10__ C. **Step-BY-STEP Explanation:** For the string to become loose, the electric force must equal the weight:
Fe=mg=0.1×10=1 NF_e = mg = 0.1 \times 10 = 1 \text{ N}
q=Fer2kQ=1×(0.1)29×109×2×10−4=5.6×10−9 Cq = \frac{F_e r^2}{k Q} = \frac{1 \times (0.1)^2}{9 \times 10^9 \times 2 \times 10^{-4}} = 5.6 \times 10^{-9} \text{ C}

All Questions in this Chapter

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