Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

A particle A of mass m and charge Q moves directly towards a fixed particle B. Which has charge Q. The speed of A is v when it is far away from B. The minimum separation between the particle is proportional to

Options:

v

v2

v-1

v-2

Correct Answer:

v-2

Explanation:

From Conservation of energy

(KE + EPE)minimum separation = (KE + EPE)far away

$\Rightarrow 0+\frac{1}{4 \pi \varepsilon_{\circ}} \frac{Q^2}{r_{\min }}=\frac{1}{2} m v^2+0$

$\Rightarrow r_{\min } \propto \frac{1}{v^2}$

Hence, (D) is correct choice.