Question

A charge 'q' is kept fixed and another charge '2q' is thrown with a velocity 'v' towards 'q' along the line joining the charges. If initial distance between the charges is very large, what will be the minimum distance between them? ​

Answer 1

Given:

A charge 'q' is kept fixed and another charge '2q' is thrown with a velocity 'v' towards 'q' along the line joining the charges. Initial separation between the charges was very large.

To find:

Minimum distance of separation between the charges?

Calculation:

In this type of questions, it is best to apply principle of Conservation of Mechanical Energy:

$\rm PE1 + KE1 = PE2 + KE2$

$\rm \implies \: \dfrac{k(q)(2q)}{ \infty} + \dfrac{1}{2} m {v}^{2} = \dfrac{k(q)(2q)}{r} + \dfrac{1}{2} m {(0)}^{2}$

$\rm \implies \:0 + \dfrac{1}{2} m {v}^{2} = \dfrac{k(q)(2q)}{r} + 0$

$\rm \implies \: \dfrac{1}{2} m {v}^{2} = \dfrac{k(q)(2q)}{r}$

$\rm \implies \: r = \dfrac{4k {q}^{2} }{m {v}^{2} }$

$\rm \implies \: r = \dfrac{4 {q}^{2} }{4\pi \epsilon_{0}m {v}^{2} }$

$\rm \implies \: r = \dfrac{ {q}^{2} }{\pi \epsilon_{0}m {v}^{2} }$

So, minimum distance is:

$\boxed{ \bf \: r = \dfrac{ {q}^{2} }{\pi \epsilon_{0}m {v}^{2} } }$