A proton is traveling horizontally to the right at 4.40×10 6 m/s . Part A Find (

Question

A proton is traveling horizontally to the right at 4.40×106 m/s .

Part A

Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.30 cm .

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Part B

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Part C

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

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Part D

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

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Part E

A proton is traveling horizontally to the right at 4.40×106 m/s .

Part A

Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.30 cm .

E = N/C

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Part B

= counterclockwise from the left direction

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Part C

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

t = s

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Part D

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

E = N/C

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Part E

= counterclockwise from the left direction

Explanation / Answer

part a:

initial speed=4.4*10^6 m/s

distance=3.3 cm=0.033 m

final speed=0

if acceleration is a,

then final velocity^2-initial velocity^2=2*a*0.033

==>a=-2.933*10^14 m/s^2

then force on proton=mass*a=-4.8987*10^(-13) N

as force due to electric field=charge*electric field

electric field=force/charge=-3.062*10^6 N/C

so magnitude=3.062*10^6 N/C

direction=0 degree with left direction

part C:

time taken=(final speed-initial speed)/acceleration

=1.5*10^(-8) seconds

part D:

if it is an electron, force =mass of electron*acceleration

=-2.669*10^(-16) N

then electric field=force/charge

=1668.143 N/C

angle with left=180 degrees

note:

values used:
mass of proton=1.67*10^(-27) kg

mass of electron=9.1*10^(-31) kg

charge on proton=1.6*10^(-19) C

charge on electron=-1.6*10^(-19) C

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