question_answer

A particle of mass 2g and charge \[E=\frac{V}{x}\]C is held at a distance of 1m from a fixed charge 1mC. If the particle is released it will be repelled. The speed of particle when it is at a distance of 10 m from the fixed charge is :

A) 90 m/s  

B)        100 m/s               

C)        45 m/s                 

D)        55 m/s

Correct Answer: A

Solution :

                                Key Idea: For a particle, change in potential energy = gain in kinetic energy. Since the charge on particles is same, hence they will be repelled from each other. Therefore, change in potential energy will be equal to the gain in kinetic energy. Change in potential energy \[G=\frac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}\] \[{{m}_{1}}\] \[{{m}_{2}}\] Kinetic energy gain is \[\therefore \] \[\text{=}\frac{\text{dimensions}\,\text{of}\,\text{force }\!\!\times\!\!\text{ (length}{{\text{)}}^{\text{2}}}}{{{\text{(dimensions}\,\,\text{of}\,\text{mass)}}^{\text{2}}}}\]  \[=\frac{[ML{{T}^{-2}}]{{[L]}^{2}}}{{{[M]}^{2}}}=[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]\]