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Take for instance the standard example with a charge in front (distance $d$ from middle) of a grounded conducting sphere of radius $a$. It is not hard to show that the relevant image charge will be $-qa/d$ at a distance $a^2/d$ from the centre.

However, is the image charge the value of the net induced charge on the sphere? Does Gauss law apply in this situation? Or do we have to take the hard route of finding the electric field from the potential, then the charge density and integrating around the entire conductor?

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In the case where the original charge is outside the sphere and the image charge is inside:

  • Consider a Gaussian surface just outside the conductor
  • In this region, the electric field due to the induced charge is exactly the the same as the electric field that would be due to the image charge
  • Therefore, the electric flux is equal to $q_{\text{in}} / \epsilon_0$ where $q_{\text{in}}$ is the image charge
  • By Gauss's law, the charge contained inside is $q_{\text{in}}$, and this is the induced charge.

Of course the crucial step is that you must draw the Gaussian surface outside the conductor. For a Gaussian surface inside the conductor the image fields and actual fields wouldn't match.

In the case where the original charge $q$ is inside the sphere, this reasoning fails, because it is impossible to draw a Gaussian surface that both lies entirely inside the sphere (for the image charge field to match) and contains the induced charges (which are on the inner surface of the sphere). But in this case the calculation is easy, we must have $q_{\text{in}} = - q$ by the standard shielding argument.

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