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. (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? (b) & (c) refer to the electron as it passes through the second set of plates on the right. (b) The electron in pan (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 it con d 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. (e) 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)
The potential between the plates, V = Ed
E is the electric field and d is the distance between the plates.
V = 2780 N/C x 5 x 10-3 m
= 13.9 V

Energy of an electron in an electric potential V is,
E = eV
= (1.6 x 10-19) x 13.9
= 22.24 x 10-19 J

This energy is manifested as the kinetic energy of electron.
E = 1/2 mv2.
Where m is the mass and v is the velocity of the electron.
22.24 x 10-19 J = 0.5 x (9.1 x 10-31) x v2.
v2 = 4.888 x 1012.
v = SQRT[4.888 x 1012]
= 2.21 x 106 m/s.

b)
The electron enters with a velocity v. The force due to electric field E is acting perpendicular to its motion. There is no force in the direction of its initial velocity.
v = 2.21 x 106 m/s.
Horizontal distance traveled by the electron, s = 1 cm
Time taken to travel 1 cm, t = s/v
= (10-2 m) / (2.21 x 106 m/s)
= 4.523 x 10-9 s

The electron will be accelerated vertically for t seconds.
Acceleration, a = F/m = eE/m
= [(1.6 x 10-19) x 740] / (9.1 x 10-31)
= 130.11 x 1012 m/s2.
Initial vertical velocity, u = 0
Distance traveled, s = ut + 1/2 at2
= 0 + 0.5 x (130.11 x 1012) x (4.523 x 10-9)2.
= 1.33 mm

So the electron will travel only 1.33 mm vertically. Since this is smaller than 5 mm, the electron will exit the plates.

c)
Initial vertical velocity, u = 0
Acceleration, a = 130.11 x 1012 m/s2.
Time, t = 4.523 x 10-9 s

Consider v as the final velocity.
v = u + at
= 0 + (130.11 x 1012) x (4.523 x 10-9)
= 0.588 x 106 m/s.

d)
Horizontal velocity, Vh = 2.21 x 106 m/s
Vertical velocity, Vv = 0.588 x 106 m/s
Net velocity, V = SQRT[(Vh)2 + (Vv)2]
= SQRT[(2.21 x 106)2 + (0.588 x 106)2]
= 2.288 x 106 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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