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Olin [163]
3 years ago
9

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

he magnitude of the electric field at a distance (a) r = 12 cm, and (b) r = 20 cm from the center of the shell? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)
Physics
1 answer:
lutik1710 [3]3 years ago
4 0

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

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A flashlight contains a battery of two cells in series, with a bulb of resistance 12 Ohms. The internal resistance of each cell
Elina [12.6K]

Answer:

1.5024

Explanation:

Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.

R = 0.26 + 0.26 + 12 = 12.52

The bulb has a voltage of 2.88 volts across it. You can get the current from that.

i = E / R

i = 2.88 / 12 =

i = 0.24 amps.

Now you can get the voltage drop across the two cells.

E = ?

R = 0.26

i = 0.24 amps

E = 0.26 * 0.24

E = 0. 0624

Finally divide the 2.88 by 2 to get 1.44

Each cell has an emf of 1.44 + 0.0624 = 1.5024

4 0
2 years ago
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8 tablespoons of water.
2 x 4 = 8

I hope this helps :)
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As an object accelerates to a speed close to the speed of light, which of the following stays the same?
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A. Speed of light (Apex) 
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How is pressure measured? question 1 options: pressure is measured as force per inch. pressure is measured as force per gram. pr
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Pressure is measured as force per unit area, which is the third option.

<h3>What is pressure?</h3>

Pressure is amount of force that is applied over a given area divided by the size of this area.

Pressure is calculated by multiplying the force applied on the object by the area covered.

Since force is measured in Newtons (N) and area is measured in m², the unit of pressure is Nm².

Therefore, pressure is measured as force per unit area.

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7 0
2 years ago
the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe
Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

          β = 10 log \frac{I}{I_{o}}

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

            I / I₀ = 10 (β / 10)

            I = I₀ 10 (β / 10)

trumpet

            I1 = 1 10⁻¹² 10 (94/10)

            I1 = 2.51 10⁻³ / cm²

Thrombus

           I2 = 1 10⁻¹² 10 (107/10)

           I2 = 5.01 10-2 W / cm²

low

           I3 =1 1-12    (113/10) W/cm²

            I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

           I_total = I₁ + I₂ + I₃

           I_ttoal = 2.51 10-3 + 5.01 10-2 + ​​1.995 10-1

           I_total = 0.00251 + 0.0501 + 0.1995

           I_total = 0.25211 W / cm²

let's bring this amount to the SI system

         β = 10 log (0.25211 / 1 10⁻¹²)

           β = 114 db

7 0
3 years ago
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