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Ede4ka [16]
3 years ago
9

Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr

om the center of the shells. The outer surface of the larger shell has aradius of 3.75 m. If the inner shell contains an excess charge of -5.30 μC, find the amount of chargeon the outer surface of the larger shell.
Physics
1 answer:
Korolek [52]3 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

 The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

Where,

\epsilon_0 = permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

E= 49.000N/C

r= 4.1m

\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}

Replacing we have,

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}

q= 9.16*10^{-5} C

q= 91.6\mu C

Therefore the amoun of charge on the outer surface of the larger shell is 91.6 \mu C

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Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

Where:

I: is the current = 4.50 A

n: is the number of electrons

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A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

4 0
2 years ago
Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer dista
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Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s)  =  344 meters

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350 meters is a longer distance than 344 meters.

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adell [148]

Answer:

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Explanation:

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From the diagram below;

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Also by using Gauss Law ;

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Combination equation 1 and 2 together ; we have

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Explanation:

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