a particle of mass m and charge q is subjected to combined action of gravity and a uniform horizontal electric field of strength E. it is projected from ground with speed v in the vertical plane parallel to the field at an angle N to the horizontal . what is the maximum distance the object can travel horizontally before striking ground.

Here, the horizontal component of initial velocity will be vcosq and vertical component of velocity will be vsinq.

The horizontal distance travelled in time t will be

x = (vcosq)t – ½ gt²

The vertical distance travelled in time t will be

x = (vsinq)t – ½ gt²

so, we have

x= (vcosq)(2vsinq/g) + ½ a(2vsinq/g)²

and for maximum horizontal range, we differentiate (1) w.r.t q

dx/dq =cos2q+a/g sin2q =0

(or)

Tan2q = –g/a = –g/QE/m = –mg/QE

Thus, the maximum distance the object can travel horizontally before striking ground is

– mg / QE

Thanks & Regards,

Vasantha Sivaraj,

askIITians faculty