Two recangular plates A and B placed at a distance 2a apart, are connected to a battery to produce an electric field. There are insulators between plantes C and other two plates. A magnetic field exists along z-axis. A charged particle of mass m and charge q passes through a hole at the middle of the plate A with velocity v and strikes at Q which is the middle of the bottom edge of plate B after passing through a hole inplate C. If `E = mv^(2)//qa`, what will be the speed of the particle at Q?

Options

A. `vsqrt(2)`
B. `2v`
C. `vsqrt(5)`
D. `vsqrt(3)`

Answer

C

Explanation

Work done `=qE.2a=` change in Ke
`q*(mv^2)/(qa)*2a=1/2 mv_(Q)^(2)-1/2 mv^(2)`
`4v^(2)=v_(Q)^(2)-v^(2) implies v_(Q)=v sqrt(5)`.