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  • Need clarity, kindly explain! - Electrostatics - JEE Main-2

A charged particle q is shot towards another charged particle  Q which is fixed, with a speed \nu it approaches  Q  upto a closest distance  r  and then  returns  ,If q were given a speed 2\nu the closest distances of approach would be

  • Option 1)

    r

  • Option 2)

    2r

  • Option 3)

    r/2

  • Option 4)

    r/4

 

Answers (2)

best_answer

As we learnt in

Potential Energy Of System Of two Charge -

U=\frac{kQ_{1}Q_{2}}{r}  \left ( S.I \right )

U= \frac{Q_{1}Q_{2}}{r}  \left ( C.G.S \right ) 

 

- wherein

K=\frac{1}{4\pi \epsilon _{0}}

 

 

 

Energy is conserved in the phenomenon

Initially,\frac{1}{2}m\nu ^{2}= \frac{kqQ}{r}\cdots \cdots \cdots \cdots (i)

Finally,\frac{1}{2}m\left ( 2\nu ^{2} \right )^{2}= \frac{kqQ}{r_{1}}\cdots \cdots \cdots \cdots (ii)

\therefore \: \: \: From\: \: (i)\: \: and\: \: (ii)

\frac{1}{4}= \frac{r_{1}}{r}\Rightarrow r_{1}= \frac{r}{4}


Option 1)

r

Incorrect

Option 2)

2r

Incorrect

Option 3)

r/2

Incorrect

Option 4)

r/4

Correct

Posted by

Aadil

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