Question

A charged particle q is shot towards another charged particle (from infinity) Q which is fixed, with a speed v. It approaches Q up to the closest distance r and then returns. If q were given a speed 2v, the closest distances of approach would be:

• A. $r$
• B. $2r$
• C. $\frac{r}{2}$
• D. $\frac{r}{4}$

Answer 1

The correct option is D $\frac{r}{4}$

When charge q is shot with velocity V:
Let r be the distance of closest approach. A/c to the law of conservation of K.E. of charge q = Electrostatic energy of charges q and Q
or, $\frac{1}{2}m{v}^2=\frac{1}{4\pi {ϵ}_0}\frac{Qq}{r}\dots \dots \left(i\right)$
When charge q is shot with velocity 2v:
Let r' be the distance of closest approach. In this case;
$\frac{1}{2}m(2v{)}^2=\frac{1}{4\pi {ϵ}_0}\frac{Qq}{r^{{}^{\prime }}}or,4\times \left(\frac{1}{2}m{v}^2\right)=\frac{1}{4\pi {ϵ}_0}\frac{Qq}{r^{{}^{\prime }}}\dots \dots (ii)$
From (i) and (ii), we have
$4\times \frac{1}{4\pi {ϵ}_0}\frac{Qq}{r}=\frac{1}{4\pi {ϵ}_0}\frac{Qq}{r^{{}^{\prime }}}$
or, $r^{{}^{\prime }}=\frac{r}{4}$