A proton is traveling horizontally to the right at 4.20x10 m/s Part A Find (a)th

Question

A proton is traveling horizontally to the right at 4.20x10 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 3.20 cm N/C Part B counterclockwise trom the left drecton Part C How much time does it take the proton to stop after entering the field? Part D What minimum field (a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)? N/C Submit Part E counnerclockwise trom the left direcion

Explanation / Answer

a)

Acceleration,

a = F / m= q E / m= 1.6 * 10-19 * E / 1.67* 10-27

a = 9.58 *107 E

From the equations of motion,

V2 - Vo2 = 2 a s

0 - (4.2 *106 )2 = 2 * 9.58* 107 E * 0.032

E = - 2.88* 106 N/C

Magnitude of electric field

|E|= 2.88* 106 N/C

b)

Direction

o=0o Counter clockwise from the left direction

c)

t=2d/V =2*0.032/(4.2*106)

t=1.52*10-8 s

(d)

Acceleration,

a = F / m= q E / m= -1.6 *10-19 * E / 9.11*10-31

a =-1.7563 *1011E

From the equations of motion,

V2 - Vo2 = 2 a s

0 - (4.2* 106 )2 =- 2 * 1.7563*1011 E * 0.032

E=1569.33 N/C

Magnitude of electric field

|E|=1569.33 N/C

e)

Direction horizontally towards right ,so

o=1800

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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