A) \[t=\frac{2m{{v}_{0}}}{qE}\]
B) \[t=\frac{2Bq}{m{{v}_{0}}}\]
C) \[t=\frac{\sqrt{3}Bq}{m{{v}_{0}}}\]
D) \[t=\frac{\sqrt{3}m{{v}_{0}}}{qE}\]
Correct Answer: D
Solution :
[d] Electric force on the particle, \[F=Eq\], and displacement\[s=\frac{1}{2}a{{t}^{2}}=\frac{1}{2}\left( \frac{Eq}{m} \right){{t}^{2}}\]. Now, \[W=\Delta K\], or \[Fs=\frac{1}{2}m(v_{f}^{2}-v_{i}^{2})\] or \[Eq\times \frac{1}{2}\left( \frac{Eq}{m} \right){{t}^{2}}\] \[=\frac{1}{2}m[{{(2{{v}_{0}})}^{2}}-v_{0}^{2}]\] \[\therefore \,\,\,\,t=\frac{\sqrt{3}m{{v}_{0}}}{qE}\]
You need to login to perform this action.
You will be redirected in
3 sec
