A proton is traveling horizontally to the right at 4.70×106 m/s . a. Find (a)the

Question

A proton is traveling horizontally to the right at 4.70×106 m/s . 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 3.40 cm . b. = 0 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)? e. counterclockwise from the left direction

Explanation / Answer

Given,

v = 4.7 x 10^6 m/s ;

a)we know that,

F = qE = ma

a = qE/m

a = 1.6 x 10^-19 x E/1.67 x 10^-27 = 9.58 x 10^7 E

We know from eqn of motion

v^2 = u^2 + 2 a S

v = 0 ; u = 4.7 x 10^6 ; a = 9.58 x 10^7 E ; S = 3.4 cm = 0.034 m

0^2 = (4.7 x 10^6)^2 - 2 x 0.034 x 9.58 x 10^7 E

2.21 x 10^13 = 6.51 x 10^6 E

E = 2.21 x 10^13/6.51 x 10^6 = 3.395 x 10^6 N/C

Hence, E = 3.4 x 10^6 N/C

Directed towards left

b)theta = 0

c)v = u + at

a = 9.58 x 10^7 x 3.4 x 10^6 = 3.26 x 10^14 m/s^2

t = u/a = 4.7 x 10^6/(3.26 x 10^14) = 1.44 x 10^-8 s

Hence, t = 1.44 x 10^-8 s

d)a = qE/m

a = 1.6 x 10^-19 x E/9.1 x 10^-31 = 1.76 x 10^11 E

We know from eqn of motion

v^2 = u^2 + 2 a S

v = 0 ; u = 4.7 x 10^6 ; a = 1.76 x 10^11 E ; S = 3.4 cm = 0.034 m

0^2 = (4.7 x 10^6)^2 - 2 x 0.034 x 1.76 x 10^11 E

2.21 x 10^13 = 1.19 x 10^10 E

E = 2.21 x 10^13/1.19 x 10^10 = 1.86 x 10^3 N/C

Hence, E = 1.86 x 10^3 N/C

Directed towards right

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.