| Directions: (16-20) |
| Motion of Charged particle in Uniform Electric Field |
| When a charged particle is placed in an electric field, it experiences an electrical force. If this is the only force on the particle, it must be the net force. The net force will cause the particle to accelerate according to Newton's second law. So |
| \[{{\overrightarrow{F}}_{e}}=q\overrightarrow{E}=m\overrightarrow{a}\] |
 |
| If \[\overrightarrow{E}\]is uniform, then \[\overrightarrow{a}\] is constant and\[\overrightarrow{a}=q\,\overrightarrow{E}/m\]. |
| If the particle has a positive charge, its acceleration is in the direction of the field. If the particle has a negative charge, its acceleration is in the direction opposite to the electric field. Since the acceleration is constant, the kinematic equations can be used. |
A) \[t=\sqrt{\frac{2hm}{eE}}\]
B) \[t=\frac{2hm}{eE}\]
C) \[t=\sqrt{\frac{2eE}{hm}}\]
D) \[t=\frac{2eE}{hm}\]
Correct Answer: A
Solution :
From Newton's law \[F=m\,\overrightarrow{a}\] Or \[qE=m\,\overrightarrow{a}\Rightarrow a=\frac{qE}{m}=\frac{eE}{m}\] Using, \[s=ut+\frac{1}{2}a{{t}^{2}}\] \[\therefore \,\,\,\,\,h=0+\frac{1}{2}\times \frac{eE}{m}{{t}^{2}}\Rightarrow t=\sqrt{\frac{2hm}{eE}}\]
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