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gizmo_the_mogwai [7]
3 years ago
8

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's int

erior, which statement is correct? A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct? The interior field points in a direction perpendicular to the exterior field. The interior field points in a direction opposite to the exterior field. There is no electric field on the interior of the conducting sphere. The interior field points in a direction parallel to the exterior field.
Physics
2 answers:
Pepsi [2]3 years ago
7 0

There is no electric field on the interior of the conducting sphere..

Gala2k [10]3 years ago
5 0

Answer:

There is no electric field on the interior of the conducting sphere.

Explanation:

When conducting sphere is placed in external electric field then as we know that all conductors will have large amount of free charge present inside the conductor, this free charge will experience force due to this external electric field and start rearrange itself.

This rearrangement of charge is known as induction in conductors

This rearrangement will continue takes place till the electric field due to this rearrangement of charge will counterbalance the external electric field.

So at the steady state position the net electric field inside all conductors must be zero

so here correct answer would be

There is no electric field on the interior of the conducting sphere.

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4 years ago
One differnce between magnetic poles and elcyrical charges is that​
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Answer:

In the electric field, the like charges repel each other, and the unlike charges attract each other, whereas in a magnetic field the like poles repel each other and the unlike poles attract each other.

Explanation:

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2. Explain brightness of light using the wave model of light.
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Read 2 more answers
You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul
photoshop1234 [79]
1) First of all, let's find the resistance of the wire by using Ohm's law:
V=IR
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega

2) Now that we have the value of the resistance, we can find the resistivity of the wire \rho by using the following relationship:
\rho =  \frac{RA}{L}
Where A is the cross-sectional area of the wire and L its length.
We already have its length L=2.90 m, while we need to calculate the area A starting from the radius:
A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2

And now we can find the resistivity:
\rho =  \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}=  1.48 \cdot 10^{-7}\Omega \cdot m
7 0
4 years ago
A 0.500-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.800 mm
Genrish500 [490]

Answer:

5.95 A

Explanation:

From the question

R = ρL/A..................... Equation 1

Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.

Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m

Substitute these values into equation 1

R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶

R = 0.084 Ω.

Finally, using Ohm law,

V = IR

Where V = Voltage, I = current

Make I the subject of the equation

I = V/R............... Equation 2

I = 0.5/0.084

I = 5.95 A

4 0
3 years ago
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