A proton is projected in the positive x direction into a region of uniform elect

Question

A proton is projected in the positive x direction into a region of uniform electric field e = (-6.30 - 10^5) T N/C at t = 0. The proton travels 8.00 cm as it comes to rest. Determine the acceleration of the proton. Determine the initial speed of the proton. Determine the time interval over which the proton comes to rest. A proton moves at 4.60 Times 10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.00 Times 10^3 N/C. Ignore any gravitational effects. Find the time interval required for the proton to travel 4.00 cm horizontally. Find its vertical displacement during the time interval in which it travels 4.00 cm horizontally. (Indicate direction with the sign of your answer.) Find the horizontal and vertical components of its velocity after it has traveled 4.00 cm horizontally.

Explanation / Answer

(a)
Charge on Proton q = 1.6*10^-19 C
E = 6.3 * 10^5 N/C

Force on proton = q*E
F = 1.6*10^-19 * 6.3 * 10^5
F = 1.008 * 10^-13 N

Force = m*a
a = F/m
a = (1.008 * 10^-13) / (1.67 * 10^-27) m/s^2
a = 6.035 * 10^13 m/s^2

(b)
Initial Speed = u
Final Speed = 0
Distance = 0.08 m

V^2 = u^2 - 2*a*s
0 = u^2 - 2* 6.035 * 10^13 * 0.08
u = 3.11 * 10^6 m/s

(c)
V= u - a*t
0 = u - a*t
t = u/a
t = 3.11 * 10^6/  6.035 * 10^13
t = 5.15 * 10^-8 s

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