# A proton moving with a speed of vo = 3.0 * 10^4 m/s 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

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 $\frac{F}{m} = \frac{E q}{m}$ 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 $\frac{2 u \sin \theta}{g}$ where,$u$ is the velocity of projection and $\theta$ is the angle of projection.

Here,replace $g$ with $\frac{E q}{m}$

So,the time to comeback to the horizontal plane is $T = \frac{2 u \sin \theta}{\frac{E q}{m}}$

Now,putting $u = 3 \cdot {10}^{4} , \theta = {30}^{\circ} , E = 400 , q = 1.6 \cdot {10}^{-} 19 , m = 1.67 \cdot {10}^{-} 27$

We get,$T = 0.78 \cdot {10}^{-} 6 = 7.8 \cdot {10}^{-} 7 s$