Answer :
Answer:
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
Answer:
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
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