Question

The electric potentials at two points A and B are \(+60\text{ V}\) and \(-30\text{ V}\) respectively. If a particle of mass \(20\,\mu\text{g}\) and charge \(+10\,\mu\text{C}\) is released from rest at point A, then the velocity with which the particle reaches point B is:

• A. \(450\text{ ms}^{-1}\)
• B. \(150\text{ ms}^{-1}\)
• C. \(300\text{ ms}^{-1}\)
• D. \(600\text{ ms}^{-1}\)

Hint:

For a charged particle moving between two potentials, directly use \(q\Delta V=\frac12 mv^2\) whenever it starts from rest.

Answer 1

The Correct Option is C

Concept: Loss of electric potential energy equals gain in kinetic energy. \[ q(V_A-V_B)=\frac12 mv^2 \]

Step 1: Calculate potential difference. \[ V_A=60\text{ V} \] \[ V_B=-30\text{ V} \] \[ \Delta V=60-(-30) \] \[ =90\text{ V} \]

Step 2: Substitute all quantities. \[ q=10\mu C =10^{-5}C \] \[ m=20\mu g =2\times10^{-8}kg \] Using \[ q\Delta V=\frac12 mv^2 \] \[ 10^{-5}\times90 = \frac12(2\times10^{-8})v^2 \] \[ 9\times10^{-4} = 10^{-8}v^2 \] \[ v^2=9\times10^4 \] \[ v=300\text{ ms}^{-1} \] \[ \boxed{300\text{ ms}^{-1}} \]