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Safari File Edit View History Bookmarks Develop Window Help 67% T- 1 Thu Jul 13 2:49:26 AM Q E 'A session.masteringphysics.com/myct/itemView?assignmentProblemiD-80396162 HW2 Exercise 21.25 Resources « previous | 1 of 13 next » Exercise 21.25 Part A A proton is traveling horizontally to the right at 5.00x106 Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.10 cm N/C Submit My Answers Give Up Part B counterclockwise from the left direction Submit My Answers Give Up Part C How much time does it take the proton to stop after entering the field? Submit My Answers Give Up

Explanation / Answer

Part A:

Here, the proton is traveling horizontally to the right.

Initial velocity = u = 5.00×10^6 m/s

And the proton uniformly comes to rest ,therefore v = 0

Distance = s = 3.10 cm =3.1*10^-2 m

Now, suppose the magnitude of the weakest electric field = E

Acceleration = a = u^2/2s=qE /m

E =mu^2/2sq

=> E =1.6726*10^-27 *( 5.0×10^6 )^2 /2*3.1*10^-2*1.6022*10^-19

=> E = Magnitude of the weakest electric field = 4.209 x 10^6 N/C

Part B:

The angle counterclockwise from the left direction is 0 (ZERO).

Part C:

time =t = sqrt [2s/a]

Acceleration = a = u^2/2s=[ 5.0×10^6] ^2 /2*3.1*10^-2 = 4.032 x 10^14 m/s^2

So, time =t = sqrt [2s/a] =sqrt [(2*3.1*10^-2) / 3.43*10^14] =1.34 x 10^-8 s

Part D:

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

E =mu^2/2sq

E =9.1095*10^-31 *( 5.0×10^6 )^2 /2*3.1*10^-2*1.6022*10^-19

=> E = 2292.6 N/C

Part E:

The requisite angle is 180 degree counterclockwise from the left direction.

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