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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.

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.

Grade:12

2 Answers

Vasantha Kumari
askIITians Faculty 38 Points
7 years ago

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

181-1448_Max_height.JPG
ria
35 Points
7 years ago
but the ans is wrong....i did it in the same way

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