AIIMS2017Physics-Electrostatics

AIIMS 2017 Physics Coulomb's Law and Friction MCQ Question

Type: MCQ-conceptual-Medium-Class 12

A particle having a charge 10 mC is held fixed on a horizontal surface. A block of mass 80 g and having charge stays in equilibrium on the surface at the distance of 3 cm from the first charge. The coefficient of friction between the surface and the block is 0.5. Find the range within the charge on the block may lie

A

−4×10⁻⁹ C to 4×10⁻⁹ C

B

−2×10⁻⁹ C to 2×10⁻⁹ C

C

−4×10⁻¹⁹ C to 4×10⁻¹⁹ C

D

−2×10⁻¹⁹ C to 2×10⁻¹⁹ C

Correct Answer

Option A

Detailed Explanation

To maintain equilibrium, the electrostatic force between the fixed charge (10 mC = 10 × 10⁻³ C) and the block must balance the maximum static friction force. The maximum static friction FfF_f is given by Ff=μmgF_f = \mu m g, where μ=0.5\mu = 0.5, m=0.08kgm = 0.08 \, \text{kg}, and g9.8m/s2g \approx 9.8 \, \text{m/s}², resulting in Ff0.4NF_f \approx 0.4 \, \text{N}. Using Coulomb's law, Fe=kq1q2r2F_e = k \frac{|q_1 q_2|}{r^2}, we find that the charge on the block q2q_2 must lie within the range of approximately 4×109C-4 \times 10^{-9} \, \text{C} to 4×109C4 \times 10^{-9} \, \text{C} to maintain equilibrium, confirming option A. Other options are incorrect as they suggest ranges that do not satisfy the balance of forces based on the given parameters.

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