Two identical point charges are moving in gravity free space, when they are 60 cm apart; their velocity vectors are equal in modulus and make angles of 45° from the line joining them as shown in the figure. If at this instant, their total kinetic energy is equal to their potential energy, what will be the distance of closest approach between them?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

40 cm

20 cm

30 cm

45 cm

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

At $r_{\min}, \frac{d r}{d t} = 0$ i.e., there wil be no components of velocities along line joining them.
$K E_{i} = \frac{1}{2} m v_{0}^{2} \times 2$

$K E_{f} = \frac{1}{2} m {(\frac{v_{0}}{\sqrt{2}})}^{2} \times 2$
$= \frac{K E_{i}}{2}$
$U_{i} + K E_{i} = U_{f} + K E_{f}$
$U_{i} + U_{i} = U_{f} + \frac{K E_{i}}{2}$
$= U_{f} + \frac{U_{i}}{2}$
$∴ U_{f} = \frac{3}{2} U_{i}$
$\frac{K q^{2}}{r} = \frac{3}{2} . \frac{K q^{2}}{60} \Rightarrow r = 40 c m$