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Lemur [1.5K]
3 years ago
8

Shadow and eclipses result from​

Physics
2 answers:
Basile [38]3 years ago
7 0

\huge\mathsf{\red{\underline{\underline{Answer}}}}

{\green{\dashrightarrow}}When an opaque obstacle is placed between a source of light and a screen, a shadow of the obstacle is formed on the screen. The kind of shadow depends on the size of the source of light. In other words, the earth casts its shadow on the moon. The solar eclipse occurs when the moon comes between the sun and the earth.

poizon [28]3 years ago
5 0
Lunar eclipses occur whenever the full moon passes into the shadow of the earth. Solar eclipses occur when the new moon slips in front of the sun and blocks it from view. The moon's shadow then falls upon the earth.
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A solid ball of radius rb has a uniform charge density rho.
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Note: question B is incomplete.

Complete Question

A solid ball of radius rb has a uniform charge density ρ.

a.  What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball?  Express your answer in terms of ρ, rb, r, and ϵ0.

b.   What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball?  Express your answer in terms of ρ, r, rb, and ϵ0.

c.   Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?

1. E(0) = 0.

2. E(rb) = 0

3. lim E(r) = 0.

4. The maximum electric field occurs when r = 0.

5. The maximum electric field occurs when r = rb.

6. The maximum electric field occurs as r to infinity.

Answer:

a) the magnitude of E(r)= ρr³/3 ε₀r²

b) the magnitude at distance r from the centre E(r)= ρr/3 ε ₀ ( if r<rb)

c) statements 1(E(0) = 0), 3(E(0) = 0) and 5(The maximum electric field occurs when r = rb.) are true

Explanation:

given

charge density = ρ ,  ε

Volume of sphere , V = (⁴/₃)πr³

a) charge density = charge/volume

ρ = q ÷ V

make charge the subject of the formula

∴q = ρ × V=  ρ× (⁴/₃)πr³

where r³ = rb³(at distance rb³)

recall

E= q/4πε₀r²

E=  ρ × (⁴/₃)πrb³/4πε₀r²

∴E(r)= ρrb³/3 ε ₀r²

(b)  The Gaussian surface is inside the ball, therefore, surface only encloses a portion of ball's charge .

The net charge enclosed by the Gaussian surface is different to the of net charge enclosed in (a)

Recall

E= q/4πε₀r²

V= (⁴/₃)πr³

E=  ρ × (⁴/₃)πr³/4πε₀r²

∴E(r)= ρr/3 ε₀

(c)  E(0)= 0

limr-----∝

E(r)= 0

The maximum electric field occurs when r=rb.

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