Question

A proton is traveling horizontally to the right at 5.00×10⁶ m/s

Part A, B:

Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.40 cm .

Part C:

How much time does it take the proton to stop after entering the field?

Part D, E:

What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?

Answer 1

a) A proton is traveling horizontally to the right,

Initial velocity = u = 5.00×10⁶ m/s

the proton uniformly comes to rest ,therefore v = zero

Distance = s = 3.40 cm =3.4*10^-2 m

the magnitude of the weakest electric field = E

Acceleration = a = u²/2s=qE /m

E =mu²/2sq  

E =1.6726*10^-27 *( 5.00×10⁶ )² /2*(3.4*10^-2)(1.6022*10^-19 )

The magnitude of the weakest electric field

E =3.7958*10⁶N/C  

the direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.40 cm is horizontally to the left

part C

time =t =

Acceleration = a = u²/2s=[ 5.0×10⁶]² /2*3.4*10^-2

a =3.67*10¹⁴ m/s^2

c) time =t = =sqrt( 2*3.5*10^-2/3.67*10¹⁴=1.36*10^-8 s

d) minimum field ((a)magnitude and (b)direction) needed to stop an electron under the conditions of part (a)=E

As electron charge is negative direction of field should be horizontal to right

E =mu²/2sq

E =9.1095*10^-31 *( 5.0×10⁶)⁶ /2*3.4*10^-2*1.6022*10^-19

E =2090.992 N/C

d ) E =2.090*10³ N/C

e) the direction of the weakest electric field that can bring the electron uniformly to rest over a distance of 3.4 cm is horizontally to the right.