Question

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

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 .

a. E =___N/C

b.∘ counterclockwise from the left direction

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

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

∘ counterclockwise from the left direction

Answer 1

a) A proton is traveling horizontally to the right,

Initial velocity = u = 5 ×10^6 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 = u^2/2s=qE /m

E =mu^2/2sq

E =1.6726*10^-27 *( 5 ×10^6 )^2 /2*3.4*10^-2*1.6022*10^-19

The magnitude of the weakest electric field E =3.8385*10^6 N/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.

b) the angle counterclockwise from the left direction is zero
______________________________
c)

time =t = sq rt 2s/a

Acceleration = a = u^2/2s=[ 5×10^6] ^2 /2*3.4*10^-2
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a =3.676*10^14 m/s^2

c) time =t = sq rt 2s/a =sq rt 2*3.4*10^-2/3.676*10^14=1.36*10^-8 s
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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 =mu^2/2sq

E =9.1095*10^-31 *( 5×10^6 )^2 /2*3.4*10^-2*1.6022*10^-19

E =2091 N/C


_______________________
the direction of the weakest electric field that can bring the electron uniformly to rest over a distance of 3.40 cm is horizontally to the right.

the angle 180 degree counterclockwise from the left direction .

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