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bearhunter [10]
2 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]2 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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Answer- There are two reasons that we know quotations have been used first is the use of of name of the person who quoted it and secondly the quotation is written inside the quotation marks.

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3 years ago
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Mama L [17]
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3 years ago
When an automobile moves with constant velocity, the power developed is used to overcome the frictional forces exerted by the ai
geniusboy [140]

Answer:

The total frictional force is 358.0 newtons

Explanation:

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F=\frac{P}{v}

Knowing that 1hp is 746W then 30hp=22380W and 1 mile = 1609m then 140 mph = 225308\frac{m}{h} = 62.5 \frac{m}{s}, then:

F=\frac{22380}{62.5}=358.0 N

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