An electron of mass m and charge q is accelerated from rest in a uniform electric field of strength E. The velocity acquired by it as it travels a distance l is

Options

A. `[(2Eql)/(m)]^(1//2)`
B. `[(2Eq)/(mI)]^(1//2)`
C. `[(2Em)/(ql)]^(1//2)`
D. `[(Eq)/(mI)]^(1//2)`

Answer

a

Explanation

The magnitude of force on a charge q in an electric field is given by
F= qE
From newton s second law we have
F=ma
From Eqs (i) and (ii) we get
`a=(qE)/(m)`
since electron starts acelerating from rest therefore initial velocity (u) of electron is zero using equaiot of motion
`V^(2)-u^(2)=2aL`
or `V^(2)-0=1(qe)/(m) I or V=sqrt(2Eql)/(m)=[(2Eql)/(m)]^(1//2)`