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:
F = 236063.6N
Explanation:
Please see attachment below.
Answer:
a) according to Faraday's law
, b) creating a faster movement, placing more turns on coil
Explanation:
a) The voltage is induced in the coil by the relative movement between it and the magnet, therefore according to Faraday's law
E = - d (B A) / dt
In this case, the magnet is involved, so the value of the magnetic field varies with time, since the number of lines that pass through the loop changes with movement.
This voltage creates a current that charges the battery
b) There are several ways to increase the voltage
* creating a faster movement, can be done by the user
* placing more turns on the coil, must be done by the manufacturer
It would be a cold wave.
Ice ages occur when the temperatures are extremely cold.(long-term)
Answer:
120 m/s
Explanation:
Given:
v₀ = 0 m/s
a = 12 m/s²
t = 10 s
Find: v
v = at + v₀
v = (12 m/s²) (10 s) + 0 m/s
v = 120 m/s
