A proton moving with a speed of vo = 3.0⋅104ms is projected at an angle of 30o above a horizontal plane. If an electric field of 400 N/C is acting down, how long does it take the proton to return to the horizontal plane?

Hint: Ignore gravitational force.

Answer is D, why?

A) 3.910-6s
B) 1.710-6s
C) 5.410-7s
D) 7.810-7s
E) 8.5*10-8s

1 Answer
Jane
Mar 15, 2018

Just compare the case with a projectile motion.

Well in a projectile motion,a constant downwards force acts that is gravity,here neglecting gravity,this force is only due to replusion by electric field.

Proton being positively charged gets replused along the direction of electric field,which is directed downwards.

So,here comparing with g,the downward acceleration will be Fm=Eqm where,m is the mass,q is the charge of proton.

Now,we know total time of flight for a projectile motion is given as 2usinθg where,u is the velocity of projection and θ is the angle of projection.

Here,replace g with Eqm

So,the time to comeback to the horizontal plane is T=2usinθEqm

Now,putting u=3⋅104,θ=30∘,E=400,q=1.6⋅10−19,m=1.67⋅10−27

We get,T=0.78⋅10−6=7.8⋅10−7s

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