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3 years ago

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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

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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From the question we are told that

The mass of the quarterback is $m_Q = 80 \ kg$

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Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball

So

$m_Q v_{Qx} = m_B v_ {Bx}$

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$v_q = \frac{0.43 * 15}{80}$

$v_q = 0.08 \ m/s$

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I hope it helps you!

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