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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.
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
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
but the ans is wrong....i did it in the same way
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