A particle of mass 0.000113 g and charge 71 mC moves in a region of space where

Question

A particle of mass 0.000113 g and charge
71 mC moves in a region of space where the
electric field is uniform and is 3.9 N/C in the
x direction and zero in the y and z direction.
If the initial velocity of the particle is given
by vy = 2.9 × 10^5 m/s, vx = vz = 0, what is
the speed of the particle at 0.3 s?
Answer in units of m/s.

So far I had used A=Q/M to determine acceleration .017 C/1.13x10^-7 KG , then plugged that into Vf=Vo+at Vf=2.9x10^5 m/s + Ax(.3)s ...I keep getting the wrong answer, does it have something to do with needing an x and y componant to the velocity and acceleration? please help me figure this out

Explanation / Answer

mass m = 0.000113 g charge q = 71 mC = 71*10^-3 C electric field in x-direction E = 3.9 N / C accleration in x-direction a = Eq / m plug the values we get a = 2450.44 m / s^ 2 Initial velocity in x-direction vx = 0 time t= 0.3 s velocity after 0.3 s in x-direction is Vx = vx + at = 735.13 m / s Initial velocity in y-direction vy = 2.9*10^5 m / s accleration a ' = 0 So, velocity after 0.3 s in y-direction Vy = vy + a' t = 2.9 * 10 ^ 5 m / s Initial velocity in z-direction vz = 0 velocity of the particle after 0.3 s in z-direction Vz = 0 Since there is no accleration in z-direction So, speed of the particle after 0.3 s V = sqrt[Vx^ 2+ Vy^2 + Vz^ 2] = 290*10^3 m / s

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