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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

A positively charged particle +q is projected with speed v toward a fixed charge +Q, and rebounds after reaching a minimum distance r. What will be the new closest distance of approach if its initial velocity is doubled to 2v?

A

r/4

B

r/2

C

r/16

D

r/8

Step-by-Step Solution

  1. Conservation of Energy: According to the conservation of mechanical energy principles discussed in Class 11 Physics [2, 3], the initial kinetic energy of the particle is entirely converted into electric potential energy at the distance of closest approach (where the particle momentarily stops).
  2. Kinetic Energy Relation: Kinetic energy (KK) is given by K=12mv2K = \frac{1}{2}mv^2 [1]. Thus, K∝v2K \propto v^2.
  3. Potential Energy Relation: The electric potential energy (UU) between two charges is inversely proportional to the distance (rr) between them (U∝1/rU \propto 1/r), analogous to the gravitational potential energy discussed in the Class 11 Gravitation chapter [4].
  4. Derivation:
  • Initial state: 12mv2∝1r  ⟹  r∝1v2\frac{1}{2}mv^2 \propto \frac{1}{r} \implies r \propto \frac{1}{v^2}.
  • New state: If the velocity is doubled (v′=2vv' = 2v), the new kinetic energy becomes K′∝(2v)2=4v2K' \propto (2v)^2 = 4v^2. This is 4 times the initial kinetic energy.
  • To balance this increased energy, the potential energy must increase by a factor of 4, which means the distance rr must decrease by a factor of 4.
  • r′=r4r' = \frac{r}{4}.

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 CAPACITANCEpositivelychargedparticleprojectedtoward

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