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I don't get it, i thought the electric feild inside a conductor is always 0. I know this is a shell and not a soild conductor but i thought it applys to this case also.

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- #1

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I don't get it, i thought the electric feild inside a conductor is always 0. I know this is a shell and not a soild conductor but i thought it applys to this case also.

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In this situation there is neither a shell by itself, or a solid by itself.mr_coffee said:

I don't get it, i thought the electric feild inside a conductor is always 0. I know this is a shell and not a soild conductor but i thought it applys to this case also.

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I believe the theory is that the inside of a conductor is always neutral - that there is an equal balance of positive and negative charge. If the total charge of the conductor isn't zero, then the residual positive or negative charge lies on the surface of the conductor.

The conductor in question now is a spherical shell, so both the inside and outside surfaces are candidates for where the residual charge can be located. Since the net charge of the shell is -100e, the sum of the charge of the inner and outer surface must equal -100e.

Like you said, the net field within a conductor is zero. This means that if you take a Gaussian surface that's complete contained inside a conductor, the integral will be equal to zero. In this example, if you were to take a Gaussian surface within the shell, the total charge enclosed would be the sum of the charge from the inner surface and the ball:

[tex]\int E \cdot dA = Q_{enc}/\epsilon_0[/tex]

[tex]\int 0 \cdot dA = (Q_i+Q_b)/\epsilon_0=0[/tex]

[tex]Q_i=-Q_b[/tex]

Q_i is the charge of the inner surface of the shell and Q_b is the charge of ball. As you can see, the charge of the inner surface must be equal and opposite to the ball, i.e. +50e. Since there is a conservation of charge, the sum of the charge of the inner and outer surfaces of the shell must equal -100e, i.e. +50e + ? = -100e. That's how you get part b.

The conductor in question now is a spherical shell, so both the inside and outside surfaces are candidates for where the residual charge can be located. Since the net charge of the shell is -100e, the sum of the charge of the inner and outer surface must equal -100e.

Like you said, the net field within a conductor is zero. This means that if you take a Gaussian surface that's complete contained inside a conductor, the integral will be equal to zero. In this example, if you were to take a Gaussian surface within the shell, the total charge enclosed would be the sum of the charge from the inner surface and the ball:

[tex]\int E \cdot dA = Q_{enc}/\epsilon_0[/tex]

[tex]\int 0 \cdot dA = (Q_i+Q_b)/\epsilon_0=0[/tex]

[tex]Q_i=-Q_b[/tex]

Q_i is the charge of the inner surface of the shell and Q_b is the charge of ball. As you can see, the charge of the inner surface must be equal and opposite to the ball, i.e. +50e. Since there is a conservation of charge, the sum of the charge of the inner and outer surfaces of the shell must equal -100e, i.e. +50e + ? = -100e. That's how you get part b.

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Doc Al

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For electrostatic conditions, this means that the field within the conducting material itself is zero. It doesmr_coffee said:I don't get it, i thought the electric feild inside a conductor is always 0. I know this is a shell and not a soild conductor but i thought it applys to this case also.

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BJ

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I am picturing something along these lines:

http://www.physics247.com/members/advanced/gaussian_surfaces.php [Broken]

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thanks again!! it makes sense now!

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