Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
Answer:
To establish this relationship we must examine the potentials that these forces create. The electrical potential is described by
Ve = k q / r
The potential for strong nuclear force is
Vn (r) = - gs / 4pir exp (-mrc / h)
Where gs is the stacking constant and r the distance between the nucleons,
We can compare these potentials where the force is derived from the relationship
E = -dU / dr
F = q E
Explanation:
The isobars in the conventional series that will be needed
to complete the pressure analysis between the lowest and highest values on this
map are: 1008, 1012, 1016, 1020.
To add, an isobar is <span>a line on a map connecting points having the
same atmospheric pressure at a given time or on average over a given period.</span>
When you're in an airplane that's 7 miles up off the ground, the strength of gravity plunges to only 99.6 percent of its strength all the way down on the ground.
A big heavy person, who weighs 200 pounds down at the airport, weighs only 199 pounds 4.7 ounces in a plane at the altitude of 7 miles.
When two mechanical waves that have positive displacements from the equilibrium position meet and coincide, a constructive interference occurs.
Option A
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Explanation:</u></h3>
Considering the principle of superposition of waves; the resultant amplitude of an output wave due to interference of two or more waves at any point is given by individual addition of their amplitudes at that point. Two waves with positive displacements refer to the fact that crest of the both the waves are on the same side of displacement axis, either both are positive or both are negative, similarly with their troughs.
If such two waves with their crest on crest meet at any point, by superposition principle. their individual amplitude gets added up and hence the resultant wave after interference is greater in amplitude that both the individual waves. This is termed as a constructive interference. Destructive interference on the other hand is a condition when one of the two waves has a positive displacement and other has a negative displacement (a condition of one’s crest on other’s trough); resulting in amplitude subtraction.
