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Consider suspending a charge inside a conductive sphere: enter image description here

The answers to this question state that there will be a charge of +Q induced on the outside of the sphere (and a charge of -Q on the inside).

I don't see why. Their argument goes as the following: due to Gauss' law, the total flux must be proportional to the total charge inside. Thus if we induce +Q and -Q, then the total sum of charge is: $$+Q_{charge} -Q_{inner sphere} +Q_{outer sphere} = +Q$$

Thus it satisfies Gauss' law. However, I really don't see why it couldn't be: $$+Q_{charge} -0.5Q_{inner sphere} + 0.5Q_{outer sphere} = +Q$$

The total flux output would still be equal to the +Q charge inside.

Can someone explain to me why the charge on the outside of the blue sphere is equal to +Q and not anything else?

Ziv Landau
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John Hon
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1 Answers1

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Just use Gauss' law one more time with Gauss spherical surface that go through the conductive sphere (i.e. spherical surface with radius that bigger than the inner radius of the conductive sphere and smaller than the outer radius of the conductive sphere and smaller than): using the fact that the electric field inside a conductive is 0 (and so does the electric flux), we get $Q_{innersphere} + Q_{charge}$ = 0. Hence, $Q_{innersphere}= -Q_{charge}$

Ziv Landau
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