In the figure, the left set of parallel plates produces a uniform electric field

Question

In the figure, the left set of parallel plates produces a uniform electric field of 2780 N/C. An electron begins at rest from the left negative plate (the plates are not drawn to scale). The right set of plates has a field of 740 N/C. Part (a) refers to the left set of parallel plates on the left. (a) How fast is the electron moving when it reaches the positive plate, 5 mm away? Parts (b) & (c) refer to the electron as it passes through the second set of plates on the right. (b) The electron in part (a) exits the first set of parallel plates through a small hole and enters the second plate, this time horizontally as shown in the figure. The second set of plates has an Electric field of 740 N/C, and a length of 1 cm (Assume the field for this second pair of plates is still all in the vertical direction and there are no edge effects distorting the field). Does the electron exit the second set of plates or does it slam into the top plate? Provide a quantitative argument. (c) What is the vertical speed of the electron as it either exits the second set of parallel plates or slams into the top? (d) What is the total velocity of the electron as it exits the second plate (or slams into the top)?

Explanation / Answer

(A) Work done by field on electron = - q (deltaV )

( and work done = change in KE )

m v^2 /2 - 0 = - (-1.6 x 10^-19) (2780)

(m = 9.109 x 10^-31 kg)

v = 3.125 x 10^7 m/s

(b) now field is in vertical direction. so horizontal velocity will not change.

time taken to cross the plates = (1 x 10^-2 m) / (3.125 x 10^7 m/s)

= 3.2 x 10^-10 sec

Fe = q E

a = q E / m = (1.6 x 10^-19) (740) / (9.109 x 10^-31) = 1.30 x 10^14 m/s^2

y = v0 t + a t^2 /2 in vertical,

y = 0 + (1.30 x 10^14) (3.2 x 10^-10)^2 /2

y = 6.655 x 10^-6 m

y is less than 5mm.

so it will not hit the plate.

(c) vy = v0 + a t

vy = 0 + (1.30 x 10^14) (3.2 x 10^-10) = 41600 m/s

(d) total velocity = sqrt((41600)^2 + (3.125 x 10^7)^2)

= 3.125 x 10^7 m/s

direction = tan^-1(41600 / (3.125 x 10^7)) = 0.07 deg

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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