A) \[q\,E\,{{y}^{2}}\]
B) \[q\,{{E}^{2}}\,y\]
C) \[q\,E\,y\]
D) \[{{q}^{2}}E\,y\]
Correct Answer: C
Solution :
Key Idea: Kinetic energy obtained by the particle is equal to the work done in moving a distance y. Electric force on charged particle F =qE Kinetic energy attained by particle = work done = force x displacement \[=qE\times y\] Alternative: Force on charged particle in a uniform electric field is \[F=ma=Eq\] or \[a=\frac{Eq}{m}\] ?(i) From the equation of motion, we have \[{{v}^{2}}={{u}^{2}}+2ay\] \[=0+2\times \frac{Eq}{m}\times y\] \[=\frac{2Eqy}{m}\] Now kinetic energy of the particle \[K=\frac{1}{2}m{{v}^{2}}\] \[=\frac{m}{2}\times \frac{2\,E\,qy}{m}=qEy\]
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