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QPages 121-124: Q29: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q29 Solution: A particle A with a charge of is held fixed on a horizontal…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the friction force as a fixed value equal to $\mu mg$, leading to a single charge value instead of a range. You must recognize that static friction adjusts to match the electric force up to a maximum limit, requiring the inequality $F_e \le \mu mg$ to define the valid bounds for $q$.

A particle A with a charge of

2.0 \times 10^{-6} \text{ C}

is held fixed on a horizontal table. A second charged particle of mass 80 g stays in equilibrium on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and this second particle is

\mu = 0.2

. Find the range within which the charge of this second particle may lie.

Step-by-Step Verified Solution

**Answer:**

2.2 \times 10^{-8} \text{ C} \le q \le 4.4 \times 10^{-8} \text{ C}

. **Step-BY-STEP Explanation:** Friction provides the balancing force. The range is determined by

\mu mg

.

f_{max} = 0.2 \times 0.08 \times 10 = 0.16

N. Electric force must equal friction:

F_e = \frac{k Q q}{r^2}

. Solving for the range of

q

gives

2.2 \times 10^{-8}

C to

4.4 \times 10^{-8}

C .

All Questions in this Chapter

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