A particle with charge `+q` and mass m, moving under the influence of a uniform electric field `Ehtai` and a uniform magnetic field `Bhatk`, follows a trajectory from P to Q as shown in figure. The velocities at P and Q as shown in figure. The velocities at P and Q are `vhati and -2vhatj` respectively. Which of the following results are incorrect?

Options

A. `E=(3)/(2)((mv^(2))/(qa))`
B. The rate of work done by the electric field at P is `(3)/(2)((mv^(3))/(a))`
C. The rate of work done by the electric field at P is 0
D. The rate of work done by the electric fields at Q is 0

Answer

A, B, C

Explanation

In going from P to Q, increase in KE
`(1)/(2)m(2v)^(2)-(1)/(2)mv^(2)=(1)/(2)m(3v^(2))=` work done by electric field
`(3mv^(2))/(2)=Eqxx2a or E=(3)/(4)((mv^(2))/(qa))`
Rate of work done by electric force at `P=vecF_(e)v`
`=(qE).v=qv[(3)/(2)((mv^(2))/(qa))]=(3)/(4)((mv^(3))/(a))`
At Q, `vecE` is perpendicular to `vecB` and v and no work is done by either field