1.

A particle of mass m and charge q is placed at rest in a uniform electric field E and then released, the kinetic energy attained by the particle after moving a distance y will beA. `q^(2)Ey`B. `qEy`C. `aE^(2)y`D. `qEy^(2)`

Answer» Correct Answer - B
Force on charged particle in a uniform electric field is
`F=ma = Eq`
`"or "a=(Eq)/(m)" ….(i)"`
From the equation of motion, we have
`v^(2)=u^(2)+2as=0+2xx(Eq)/(m)xxy=(2Eqy)/(m)`
Now, kinetic energy of the particle
`KE=(1)/(2)mv^(2)`
`=(m)/(2)xx(2Eqy)/(m)=Eqv`


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