I\'m having a hard time answering these theree questions,please help.Ill rate if

Question

I'm having a hard time answering these theree questions,please help.Ill rate if answer is correct and clear to understand.

* A proton is traveling horizontally to the right at 4.40×10^6 m/s .

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

E=_______N/C

PART B :" Theta" counter clockwise from left direction

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

T=_____s

Explanation / Answer

a) A proton is traveling horizontally to the right,

Initial velocity = u = 4.40×10^6 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 *( 4.40×10^6 )^2 /2*3.4*10^-2*1.6022*10^-19

The magnitude of the weakest electric field E =2.97*10^6 N/C
______________________________
the direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50 cm is horizontally to the left.

b) the angle counterclockwise from the left direction is zero
______________________________
time =t = sq rt 2s/a

Acceleration = a = u^2/2s=[ 4.40×10^6] ^2 /2*3.4*10^-2

a =5.69*10^14 m/s^2

c) time =t = sq rt 2s/a =sq rt 2*3.4*10^-2/3.43*10^14=1.093*10^-8 s

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