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:
the wave represents the second harmonic.
Explanation:
Given;
length of the cord, L = 64 cm
The first harmonic of a cord fixed at both ends is given as;
The wavelength of a standing wave with two antinodes is calculated as follows;
L = N---> A -----> N + N ----> A -----> N
Where;
N is node
A is antinode
L = N---> A -----> N + N ----> A -----> N = λ/2 + λ/2
L = λ
The harmonic is calculated as;
Therefore, the wave represents the second harmonic.
L = λ
The major shortcoming of Rutherford's model was that it was incomplete. It did not explain how the atom's negatively charged electrons are distrubuted in the space surronding its positively charged nucleus. A form of energy that exhibits wavelike behavior as it travels through space
Answer:
C. volume of water and temperature
Explanation:
a p e x
