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Kobotan [32]
3 years ago
7

An uncharged spherical conducting shell surrounds a charge –q at the center of the shell. Then charge +3q is placed on the outsi

de of the shell.
When static equilibrium is reached, the charges on the inner and outer surfaces of the shell are respectively:

a) +q, -q
b) -q, +q
c) +q, +2q
d) +2q, +q
Physics
1 answer:
Veronika [31]3 years ago
8 0

Answer:

a) The the charges on the inner and outer surfaces of the shell are respectively +q, -q

Explanation:

Under static equilibrium inside a conductor, the total electric field, E = 0

This must be zero so that no charge will be moving since the conductor is in static equilibrium.

Also, since Electric field, E is zero, then flux through the surface will zero.

From Gauss' law, the total charge enclosed is zero.

Given –q as the  charge at the center of the shell, then the opposite charge on inner surfaces  will be +q, so that the total charge enclosed will be zero.

Since the charge is in static equilibrium, then opposite charge will be on the surface, that is –q.

Therefore, the the charges on the inner and outer surfaces of the shell are respectively +q, -q

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A machine does 500 j of work in 20 sec. What is the power of this machine?
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A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, an
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Answer:

Explanation:

Mass of 190kg

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Coefficient of kinetic friction 0.36

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W=190×9.81=1863.91N

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Fr= 0.4×1863.91

Fr=745.56N.

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Then the frictional force at that time is equal to the horizontal force

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b. Mass of asteroid is

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Asteroid velocity at a particular instant is,

U=(-1.30x10^4, 4.20x10^4, 0)m/s

Magnitude of U is

U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0

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Position of the asteroid from the centre of the earth is,

R= (6.00x10^6, 10.00x10^6, 0)m.

The magnitude of the radius is

R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2

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The new momentum is 9.66×10^7 Kgm/s

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