Question

A charged particle 'q' is shot with speed v towards another fixed charged particle Q. It approaches Q upto a closest distance r and then returns. If q were given a speed $2v$, the closest distance of approach would be.

• A. r
• B. $2r$
• C. $r/2$
• D. $r/4$

Answer 1

The correct option is D $r/4$

At closest distance r its whole KE is converted into PE.
$\therefore \frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{Q\cdot q}{r}$ $\Rightarrow r=\frac{1}{4\pi {\epsilon }_0}\frac{Q\cdot q}{m{v}^2}$
In next case, $r^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{m(2v{)}^2}$
$\Rightarrow r^{\prime }=\frac{1}{4}\left(\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Qq}{m{v}^2}\right)\Rightarrow r^{\prime }=r/4$.