Question

A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° 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?

Answer 1

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Step-by-step explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

`T = (2usin \theta)/((Eq)/(m)) =(m* 2usin \theta)/(Eq)`

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

`T = ((1.67X10^(-27)* 2*3X10^4sin 30)/(400*1.6X10^(-19))`

T = 7.83 X10⁻⁷ s