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bearhunter [10]
3 years ago
10

Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen

ter of the spherical shell, then the electric field between the two, as well as the field outside the outer boundary of the conductor, is the same as the field you would measure if the conducting shell was not there, though the charges of the conductor will redistribute themselves to ensure zero E field inside the conductor.
a. True b. False
Physics
1 answer:
Furkat [3]3 years ago
8 0

Answer:

True

Explanation:

If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell

According to Gauss law

∅ = EA =-Q/∈₀

Where ∅  is the electric flux through the gaussian surface and E is the electric field strength

If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero

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A cell phone uses a 3. 0 V battery. The circuit board it uses needs a 0. 05 A current. What size resistor is needed to generate
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The value of resistance will be 60 ohms. Resistance is the opposition force offered to the flow of electric current. The ratio of the voltage to the electric current passing through the wire.

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I is value of electric current=  0. 05 A

R is the resistance=?

According to ohm's law, the potential difference between the two ends of the conductor is equal to the product of electric current and the resistance generated.

So from the ohm law;

V= IR

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Hence the value of resistance will be 60 ohms.

To learn more about the resistor refer to the link;

brainly.com/question/1322424

3 0
2 years ago
For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate
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Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue=h_1=0.200 m

Metal volume flow  rate,Q=0.03m^3/min

Q=\frac{0.03}{60}=5\times 10^{-4}m^3/s because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

d_=0.030 m

h_2=0

A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}

A_1=7.065\times 10^{-4} m^2

v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}

v_1=0.708 m/s

Pressure at the top and bottom of the sprue is atmospheric

h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}

Substitute the values

0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}

v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264

v_2=\sqrt{4.421264}=2.1 m/s

Q=A_2v_2

5\times 10^{-4}=A_2\times 2.1

A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2

Reynolds number=\frac{v_2D\rho}{\eta}

\eta=0.004 N.s/m^2

\rho=2700 kg/m^3

Substitute the values then we get

Reynolds number=\frac{2.1\times 0.03\times 2700}{0.004}

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

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