A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a dist
1 answer:
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
=
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
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Explanation:
Given that,
Radius R= 2.00
Charge = 6.88 μC
Inner radius = 4.00 cm
Outer radius = 5.00 cm
Charge = -2.96 μC
We need to calculate the electric field
Using formula of electric field
(a). For, r = 1.00 cm
Here, r<R
So, E = 0
The electric field does not exist inside the sphere.
(b). For, r = 3.00 cm
Here, r >R
The electric field is
Put the value into the formula
The electric field outside the solid conducting sphere and the direction is towards sphere.
(c). For, r = 4.50 cm
Here, r lies between R₁ and R₂.
So, E = 0
The electric field does not exist inside the conducting material
(d). For, r = 7.00 cm
The electric field is
Put the value into the formula
The electric field outside the solid conducting sphere and direction is away of solid sphere.
Hence, This is the required solution.