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pentagon [3]
3 years ago
14

A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A

small tack with charge Q is lowered on a silk thread through the hole into the interior of the shell.
(i) What is the charge on the inner surface of the shell?
A) Q/2
B) 0
C) -Q/2
D) -Q
Physics
1 answer:
cluponka [151]3 years ago
5 0

Answer:

D) -Q

Explanation:

The charge inserted will induce -Q charge on the inner surface and + Q on the outer surface of the shell . This charge is called bound charge because it remained attached with opposite charge inserted inside.

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A solid ball of radius rb has a uniform charge density rho.
Oksana_A [137]

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.

4 0
3 years ago
A lightbulb consumed 1100J in 30 minutes. Find Power, resistance, and calculate energy in kWh
lisov135 [29]

Explanation:

30 minutes is 1800 seconds.

Power = energy / time

P = 1100 J / 1800 s

P = 0.611 W

Converting the energy from J to kWh:

1100 J × (1 Ws / 1 J) × (1 kW / 1000 W) × (1 h / 3600 s) = 3.06×10⁻⁴ kWh

To find resistance, you need to be given either the voltage or the current.

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finlep [7]

Answer:

True

Explanation:

because they can hold marine organism inside

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F e d b c a
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6 0
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Find the orbital speed v for a satellite in a circular orbit of radius R.Express the orbital speed in terms of G, M, and R.
AlekseyPX
<h2>Answer:V=\sqrt{G\frac{M}{R}}  </h2>

The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:

V=\sqrt{G\frac{M}{R}}     (1)

Where:

G is the gravity constant

M the mass of the massive body around which the satellite is orbiting

R the radius of the orbit (measured from the center of the planet to the satellite).

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>

In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.

5 0
3 years ago
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