If two particles have equal charges and are placed near one another, how
1 answer:
Answer: C) The two particles will move away from each other
Explanation:
When two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies.
This is stated by Coulomb's Law:
"The electrostatic force between two point charges
and
is proportional to the product of the charges and inversely proportional to the square of the distance
that separates them, and has the direction of the line that joins them"
Mathematically this law is written as:
Where is a proportionality constant.
Now, if and
have the same sign charge (both positive or both negative), a repulsive force will act on these charges.
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Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t = ( m + 1/2) × λ/2n
we substitute
t = ( 0 + 1/2) × 711 /2(1.21)
t = 0.5 × 711/2.42
t = 0.5 × 293.80165
t = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm