# A charge +q is ejected from infinity with velocity v towards +Q charge ,it stops at a distance of R from the +Q charge.At what distance it will stop if it is ejected with 2v velocity

In the Ist case, applying the conservation of the energy

Potential energy of change +

*q*when it stop at distance R from change +

*Q*= Initial kinetic energy of the charge +

*q*

$\frac{Qq}{4\pi {\epsilon}_{0}R}=\frac{1}{2}{m}_{q}{v}^{2}....\left(i\right)$

Consider that, when change +

*q*is projected with velocity 2

*v*is stop at distance

*R*' from charge +

*Q*.

applying the conservation of the energy, we get

$\frac{Qq}{4\pi {\epsilon}_{0}R\text{'}}=\frac{1}{2}{m}_{q}(2v{)}^{2}....(ii)$

Dividing equation (i) by (ii), we get

$R\text{'}=\frac{R}{4}$

Regards,

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