A) \[\frac{1}{2}q\,E{{y}^{2}}\]
B) \[qEy\]
C) \[qE{{y}^{2}}\]
D) \[q{{E}^{2}}y\]
Correct Answer: B
Solution :
Here, \[u=0;\,a=\frac{qE}{m};\,s=y\] Using, \[{{v}^{2}}-{{u}^{2}}=2as\Rightarrow {{v}^{2}}=2\frac{qE}{m}y\] \[\therefore \,\,\,\,K.E.=\frac{1}{2}m{{v}^{2}}=qEy\].
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