A) \[\frac{1}{2}qE{{y}^{2}}\]
B) \[qEy\]
C) \[qE{{y}^{2}}\]
D) \[q{{E}^{2}}y\]
Correct Answer: B
Solution :
(b) \[qEy\] Here, \[u=0;\,a=\frac{qE}{m};\,s=y\] Using, \[{{v}^{2}}-{{u}^{2}}=2as\Rightarrow {{v}^{2}}=2\frac{qE}{m}y\] \[\therefore \,\,\,K.E.=\frac{1}{2}m{{v}^{2}}=qEy\]You need to login to perform this action.
You will be redirected in
3 sec