A proton is traveling horizontally to the right at 4.40*10^6. a) find the magnit

ID: 1536382 • Letter: A

Question

A proton is traveling horizontally to the right at 4.40*10^6.

a) find the magnitude and direction of the weakest electric fiedl that can bring the proton uniformly to rest over a distance of 3.00 cm

b) heta answer is 0 counter clock wise from the left direct

c) how much time does it take the proton to stop after entering the field?

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

e) heta? counterclockwise from the left direction

(a)Initial speed =u= 4.40*106 m/s

Final speed = v= 0

mass of proton = m= 1.672 × 10-27kg

charge of proton= q= 1.6 * 10-19 C

distance = 0.03m

Force due to electric field = qE

From newtons laws of kinematics

v2 = u2 + 2as

0=(4.40*106)2+2 a (0.03)

a= - 322 * 1012=-3.22 * 1014m/s2

qE = ma

E= (ma)/q

=(1.672 × 10-27)(3.22 * 1014)/(1.6 * 10-19 )= 3.36 * 106N/C

(b) Direction of electric fiels from right to left . zero degrees counterclockwise from left.

t= -u/a

=- (4.40*106)/ (-3.22 * 1014)= 1.366 * 10-8 sec

(d) as electron and proton have same charge, same electic field is needed to stop the electron.

E=3.36 * 106N/C

(e) direction of electric field is in direction of motion of electron

=180 degrees counter clockwise from left direction.

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