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 is zero
D. Rate of work done both the fields at Q is zero

Answer

A, B, D

Explanation

Magnetic force does not do work. From work-energy theorem `W_(F_(e))=DeltaDE` or `(qE)(2a)=1/2m[4v^(2)-v^(2)]`
or `E=3/4((mv^(2))/(qa)):.` correct option is (A)
At P, rate of work done by electric fiedl `=vecF_(e).vecv=(qE)(v)cos 0^(@)=q(3/4 (mv^(2))/(qa))v=3/4((mv^(3))/a)`
Therefore, option (B) is also correct. Rate of work done at Q : of electric field
`=F_(e).v=(qE)(2v)cos 90^(@)=0`
And of magnetic field is always zero. Therefore option (D) is also correct. Note that `F_(e)=qEhati`