Question

"Oppositely charged parallel plates are separated by T 5.33 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? (b) What is the magnitude of the force on an electron between the plates? (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.90 mm from the positive plate?"

Answer 1

a) the magnitude of the electric field between the plates is 1.12570 × 10⁵ N/C

b)the magnitude of the force on an electron between the plates 1.80 × 10⁻¹⁴N

c)4.37 × 10⁻¹⁷ J

Step-by-step explanation:

Given that,

potential difference of 600 V

plates are separated by T 5.33 mm =5.33 * 10⁻³m

E=V/d

= 600 / 5.33 * 10⁻³

= 1.12570 × 10⁵ N/C

b) F = E*q

F = 1.12570 × 10⁵ × 1.6*10⁻¹⁹

F = 1.80 × 10⁻¹⁴N

c)Work = F*d

W = 1.80 × 10⁻¹⁴ × (5.33 - 2.90) × 10⁻³

W = 1.80 × 10⁻¹⁴ × 2.43 × 10⁻³

W = 4.37 × 10⁻¹⁷ J

Answer 2

Step-by-step explanation:

Given that,

Plate separation

d = 5.33mm = 0.00533m

Potential difference

V = 600V

A. Magnitude of electric field?

The magnitude of electric field is given as

E = |-∆V/∆x|

E = ∆V/∆x

E = 600/0.00533

E = 112,570.36 N/C

E = 1.126 × 10^5 N/C

B. Force on an electron between plate?

Force in an electric field is determine by using the formula

F = qE

Where q is an electron of charge

q = e = 1.609×10^-19C

F = eE

F = 1.609×10^-19 × 1.126×10^5

F = 1.81 × 10^-14 N

C. Work done to move the negative plate from 5.33mm to 2.90mm

Work done is given as

W = - F∆x Cosθ

The angle between the force and the displacement is 0°

∆x = x2 - x1

∆x = 2.90 — 5.33

∆x = -2.43mm

∆x = -0.00243m

So, W = -F∆xCosθ

W = - 1.81×10^-14 × -0.00243Cos0

NOTE: -×- =+, Cos0 =1

W = 4.398 × 10^-17 J

W ≈ 4.4 × 10^-17 J