A proton is traveling horizontally to the right at 5.00 times 10^6 m/s You may w

Question

A proton is traveling horizontally to the right at 5.00 times 10^6 m/s You may want to review for related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Find (a) the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50 cm. How much time does it take the proton to stop after entering the field? What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?

Explanation / Answer

a) let a is the acceleration of the proton.

use, v^2 - u^2 = 2*a*s

==> a = (v^2 - u^2)/(2*s)

= (0^2 - (5*10^6)^2)/(2*0.035)

= -3.57*10^14 m/s^2

|a| = 3.57*10^14 m/s^2

now use,

F = q*E

m*a = q*E

==> E = m*a/q

= 1.67*10^-27*3.57*10^14/(1.6*10^-19)

= 3.73*10^6 N/c

b) 0 degrees

c) apply, v = u + a*t

==> t = (v - u)/a

= ( 0 - 5*10^6)/(-3.57*10^14)

= 1.4*10^-8 s

d) E = m*a/q

= 9.11*10^-31*3.57*10^14/(1.6*10^-19)

= 2.03*10^3 N/c

e) direction : theta = 180 degrees

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