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Electrostatics question

2004 · Shift 0 · Q181
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Electrostatics question

2004 · Shift 0 · Q181

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A charge particle ′q′'q'′q′ is shot towards another charged particle ′Q′'Q'′Q′ which is fixed, with a speed ′v′'v'′v′. It approaches ′Q′'Q'′Q′ upto a closest distance rrr and then returns. If qqq were given a speed of ′2v′'2v'′2v′ the closest distances of approaches would be
  1. A
    r/2r/2r/2
  2. B
    2r2r2r
  3. C
    rrr
  4. D
    r/4r/4r/4
View written solutionFree

Correct answer: D

  1. Interpret the situation

Since the particle approaches the fixed charge QQQ, stops at the closest distance, and then returns, the interaction must be repulsive.

At the point of closest approach, the moving charge qqq momentarily comes to rest, so its initial kinetic energy is completely converted into electrostatic potential energy.

  1. Apply conservation of energy for the first case

If the initial speed is vvv, then

12mv2=kQqr\frac{1}{2}mv^2 = \frac{kQq}{r}21​mv2=rkQq​

where rrr is the distance of closest approach.

  1. Apply conservation of energy for the second case

If the speed is doubled to 2v2v2v, then at the new closest distance r′r'r′,

12m(2v)2=kQqr′\frac{1}{2}m(2v)^2 = \frac{kQq}{r'}21​m(2v)2=r′kQq​

Simplify the left side:

12m(4v2)=2mv2\frac{1}{2}m(4v^2) = 2mv^221​m(4v2)=2mv2

So,

2mv2=kQqr′2mv^2 = \frac{kQq}{r'}2mv2=r′kQq​
  1. Use the first equation to relate r′r'r′ and rrr

From the first case,

12mv2=kQqr\frac{1}{2}mv^2 = \frac{kQq}{r}21​mv2=rkQq​

Multiply both sides by 444:

2mv2=4kQqr2mv^2 = \frac{4kQq}{r}2mv2=r4kQq​

But from the second case,

2mv2=kQqr′2mv^2 = \frac{kQq}{r'}2mv2=r′kQq​

Therefore,

kQqr′=4kQqr\frac{kQq}{r'} = \frac{4kQq}{r}r′kQq​=r4kQq​

Cancelling kQqkQqkQq,

1r′=4r\frac{1}{r'} = \frac{4}{r}r′1​=r4​

Hence,

r′=r4r' = \frac{r}{4}r′=4r​
  1. Check options
  • A: r/2r/2r/2 ❌
  • B: 2r2r2r ❌
  • C: rrr ❌
  • D: r/4r/4r/4 ✅

So the correct answer is Option D.

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