Question

A charged particle q is fired towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q is given speed 2v, the closest distance of approach would be: [AIEEE-2004]

• A. r
• B. 2r
• C. r/2
• D. r/4

Answer 1

The correct option is D r/4

As the electrostatic force is a conservative force, we can conserve total energy.
$(K.E{)}_i+(P.E{)}_i=(K.E{)}_f+(P.E{)}_f$
Let us consider the position from which 'q' is fired as our reference point, because what only matters is the change in P.E.

with speed v:
$0+\frac{1}{2}m{v}^2=\frac{KqQ}{r}+0$
$\frac{1}{2}m{v}^2=\frac{KqQ}{r}\Rightarrow r=\frac{2KqQ}{m{v}^2}$ ...(1)

with speed 2v:
$\frac{1}{2}m(2v{)}^2+0=0+\frac{KqQ}{r^1}$

where $r^1$ is the new closest distance of approach
$\Rightarrow 2m{v}^2=\frac{KqQ}{r^1}$
$r^1=\frac{KqQ}{2m{v}^2}$
from $\left(1\right),r^1=\frac{r}{4}$