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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

The electrostatic force between the metal plates of an isolated parallel plate capacitor CC having a charge QQ and area AA is:

A

independent of the distance between the plates.

B

linearly proportional to the distance between the plates.

C

proportional to the square root of the distance between the plates.

D

inversely proportional to the distance between the plates.

Step-by-Step Solution

To find the force between the plates, we consider the electric field produced by one plate acting on the charge of the other plate.

  1. Electric Field of a Single Plate: The electric field EE due to a single large plane sheet of charge with surface charge density σ=Q/A\sigma = Q/A is given by E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0} [NCERT Class 12, Chapter 1, Application of Gauss's Law].
  2. Force Calculation: The force FF exerted on the second plate carrying charge QQ (placed in the field of the first plate) is F=Q×EF = Q \times E. F=Q×σ2ϵ0=Q×Q2Aϵ0F = Q \times \frac{\sigma}{2\epsilon_0} = Q \times \frac{Q}{2A\epsilon_0} F=Q22ϵ0AF = \frac{Q^2}{2\epsilon_0 A}
  3. Conclusion: The expression for force depends only on the charge QQ, the area AA, and the permittivity ϵ0\epsilon_0. It does not contain the separation distance dd. Therefore, the force is independent of the distance between the plates (provided the distance is small compared to the plate dimensions, maintaining the uniform field approximation).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. 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.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEelectrostaticbetweenplatesisolatedparallel

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