Answer:
a) Electric potential = 853 V
b) Electron speed at point B, if at Point A, the speed were zero = 1.732 × 10⁷ m/s
Explanation:
For an electron moving in an electric field with potential V,
Work done = qV where q is the charge on the electron
And the Work done is equal to the change in kinetic energy of the electron
qV = m(v₂² - v₁²)/2
V = m(v₂² - v₁²)/2q
q = 1.602 × 10⁻¹⁹C
m = 9.11 × 10⁻³¹ kg
v₁= 10⁷ m/s
v₂ = 2 × 10⁷ m/s
Putting these values in for the variables and solving
V = 853 V
b) If the electron started from rest,
qV = mv²/2
v = √(2qV/m) =√((2 × (1.602 × 10⁻¹⁹) × 853)/(9.11 × 10⁻³¹)) = 1.732 × 10⁷ m/s
Answer:
λ= 2455 nm
Explanation:
Constructive interference occurs when the optical path difference is equal to an integer multiple of wavelength. Let's examine the reflection on each face
On the first surface we go from a lower medium refractive index (air) to one with a higher index (film), whereby a phase change of pi (180º) is introduced. On the second surface instead of the beginning of refraction, it is from a higher index to a smaller one, so there is no phase change, with this we can write the normal interference equation.
2 t = (m + ½)
The term lann is the wavelength in the film that is related to the wavelength in the air by
= λ / n
2t = (m + ½) λ / n
λ = 2t n / (m + ½)
For the first constructive interference m = 0
λ = 2 0.448 10⁻⁶ 1.37 / (0 + ½)
λ = 2,455 10⁻⁶ m
λ = 2,455 10⁻⁶ m (10⁹ nm / 1m)
λ= 2455 nm
For m = 1
λ = 2 0.448 10-6 1.37/ (1 + ½)
λ = 0.81835 10-6 m
λ = 818 nm
We can observe that the length and where it corresponds to ultraviolet light, so there is no constructive interference for visible light.
Answer:
TRUE. The potential of a negatively charged conductor must be negative
Explanation:
Let's examine each statement
The positively charged proton moves in the direction of the electric field, the power and the electric field are related
ΔU = - E ds
- U₀ = - E ds
E = (U₀ –U_{f}) / s
To have a positive electric field the initial potential must be greater than the final potential, so the proton moves from a greater potential to a smaller one.
This statement is FALSE
The second statement
The potential has the same sign as the elective charge.
This statement is TRUE