A particle of charge ` +q` and mass `m` moving under the influence of a uniform electric field ` E hat(i)` and uniform magnetic field ` B hat(k)` follows a trajectory from ` P to Q` as shown in fig. The velocities at `P and Q` are ` v hat(i)` and ` - 2v hat(j)` . which of the following statement(s) is/are correct ?

Options

A. `E=3/4((mv^2)/(qa))`
B. Rate of work done by the electric field at P is `3/4[(mv^3)/(a)]`
C. Rate of work done by the electric field at `P=0`
D. Rate of work done by both fields at Q is zero

Answer

A, B, D

Explanation

Increase in K.E=work done by electric field
`1/2m(2v)^2-1/2mv^2=qE(2a)`
or `E=3/4((mv^2)/(qa))`
(b) Rate of work done by electric field at P
=Electric force x Velocity at P
`qExxv=q[3/4(mv^2)/(qa)]v`, `=3/4(mv^3)/(a)`
(d) At Q, force acting on particle due to electric field is zero (as `2vhatj`) is perpendicular to both `Ehati` and `Bhatk`) and hence rate of work done by both fields is zero.