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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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<h3>What is the mass of Jupitar?</h3>

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Given that;

T^2 = GMr^3/4π

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1 day = 86400 seconds

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Making M the subject of the formula;

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M = 5.8 * 10^-14 Kg

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A plant was placed in the corner of a room away from a window. What might you observe about the plant after five days?
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\texttt{Average speed = }\frac{\texttt{Distance}}{\texttt{Time}}\\\\\texttt{Average speed = }\frac{0.15}{20}=7.5\times 10^{-3}m/s\\\\\texttt{Average velocity = }\frac{\texttt{Displacement}}{\texttt{Time}}\\\\\texttt{Average velocity = }\frac{0.05}{20}=2.5\times 10^{-3}m/s

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