First slide

Potential energy

Question

A particle of mass 2g and charge $1 μC$ is held at rest on a frictionless horizontal surface at a distance of 1m from a fixed charge 1mC. If the particle is released, it will be repelled. The speed of the particle when it is at a distance of 10m from the fixed charge is :

Moderate

A

100 m/s
B

90 m/s
C

60 m/s
D

45 m/s

Solution

Gain in KE= Loss in electrostatic potential energy
$\Rightarrow \frac{1}{2} {mv}^{2} = \frac{q_{1} q_{2}}{4 {πε}_{0}} (\frac{1}{r_{1}} - \frac{1}{r_{2}}) \Rightarrow \frac{1}{2} x 2 x 10^{- 3} x v^{2} = 9 x 10^{9} x 10^{- 3} x 10^{- 6} [\frac{1}{1} - \frac{1}{10}] \Rightarrow v^{2} = 8100 \Rightarrow v = 90 m / s$

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