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NEET PHYSICSELECTRIC CHARGES AND FIELDSMedium

Question

Two equally charged, identical metal spheres A and B repel each other with a force FF. The spheres are kept fixed with a distance rr between them. A third identical, but uncharged sphere C is brought in contact with A, then brought in contact with B, and finally placed at the mid-point of the line joining A and B. The magnitude of the net electric force on C is:

A

F

B

3F/4

C

F/2

D

F/4

Step-by-Step Solution

Let the initial charge on spheres A and B be qq. The force F=14πϵ0q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}.

  1. Contact with A: When uncharged sphere C touches A, the charge qq distributes equally. Charge on A becomes qA=q/2q_A = q/2, and charge on C becomes qC=q/2q_C = q/2.
  2. Contact with B: Sphere C (now with charge q/2q/2) touches sphere B (charge qq). Total charge is q+q/2=3q/2q + q/2 = 3q/2. This splits equally, so qB=3q/4q_B = 3q/4 and qC=3q/4q_C = 3q/4.
  3. Placed at Midpoint: C is placed at distance r/2r/2 from both A and B. Force from A on C (FCAF_{CA}): Repulsive. FCA=14πϵ0(q/2)(3q/4)(r/2)2=14πϵ03q2/8r2/4=1.5FF_{CA} = \frac{1}{4\pi\epsilon_0} \frac{(q/2)(3q/4)}{(r/2)^2} = \frac{1}{4\pi\epsilon_0} \frac{3q^2/8}{r^2/4} = 1.5 F. Force from B on C (FCBF_{CB}): Repulsive. FCB=14πϵ0(3q/4)(3q/4)(r/2)2=14πϵ09q2/16r2/4=2.25FF_{CB} = \frac{1}{4\pi\epsilon_0} \frac{(3q/4)(3q/4)}{(r/2)^2} = \frac{1}{4\pi\epsilon_0} \frac{9q^2/16}{r^2/4} = 2.25 F.
  4. Net Force: Since forces are opposite, Fnet=FCBFCA=2.25F1.5F=0.75F=3F4F_{net} = F_{CB} - F_{CA} = 2.25F - 1.5F = 0.75F = \frac{3F}{4}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTRIC CHARGES AND FIELDS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTRIC CHARGES AND FIELDSequallychargedidenticalspheresspheres

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