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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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