Question

Hello,

Please help me with this problem.

Show all work clearly. No credit for answer only. Provide the proper units. I. A particle of mass 0.20 kg and charge 2.5 μC has an initial velocity of 20.0 m/s in the x-direction. The particle enters a uniform electric field pointed downward in the negative y-direction. If the particle's initial height above the ground is 15.0 meters, how far will the particle travel before it strikes the ground? E = 2.0×104 N/C

Answer 1

Here ,

initial velocity of particle, u = 20 i m/s

E = -2 *10^4 j N/C

let a be the acceleration is a

using second law of motion

q*E = m*a

2.5 *10^-6 * 2 *10^4 = 0.20 * a

a = 0.25 m/s^2

Now, time taken to strike ground is t

using second equation of motion in vertical direction

d = ut + 0.5 at^2

15 = 0.5 * 0.25 * t^2

t = 10.95 s

horizontal distance = t*u

horizontal distance = 10.95 * 20

horizontal distance = 219.1 m

the horizontal distance by the particle is 219.1 m