A) \[\frac{m{{u}^{2}}}{2qE}\]
B) \[\frac{mu}{qE}\]
C) \[\frac{qE}{m}\]
D) None of these
Correct Answer: A
Solution :
Particle is thrown against electric field \[\therefore \,\,\,\,\,\operatorname{Re}tardation\,\,is\,\,a=-\frac{F}{m}\] \[=-\frac{qE}{m},\,\,(\left| F \right|\,\,=\,\,q\,E)\] Now, \[{{\operatorname{v}}^{2}}- {{u}^{2}}= 2as\] \[\Rightarrow \,\,\,\,\,0-{{u}^{2}}=2\frac{(-\,qE)}{m}\,\,s\,\,\Rightarrow \,\,s=\frac{m{{u}^{2}}}{2qE}\]
You need to login to perform this action.
You will be redirected in
3 sec
