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I read the description and the explanation of the result of the experiment in my book but still I can't understand it.

Here is the process of the experiment.

A pair of concentric metallic spheres constructed, the outer one consisting of two hemispheres that could be firmly clamped together. shells of insulating material (or dielectric material, or simply dielectric) that would occupy the entire volume between the concentric spheres.

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The result is the total charge on the outer sphere was equal in magnitude to the original charge placed on the inner sphere and that this was true regardless of the dielectric material separating the two spheres

Faraday explain this by Electric displacement field but still I cant understand the result of the experiment

The outer is connected to the ground so its charge must be zero so how can those 2 sphere have the same magnitude charge at the end. If the outer isn't connected to the ground then can it get -2 charge at the end?

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  • $\begingroup$ The outer is momentarily connected to the ground. This is to remove any excess charges that might have built up. $\endgroup$ – AV23 May 12 '15 at 20:42
  • $\begingroup$ Could you clarify why you think momentarily is important? Even if you left the ground connected for relatively long time, the result of the experiment will be the same, I would think. $\endgroup$ – ApproximatelyTrue May 13 '15 at 0:27
  • $\begingroup$ @ApproximatelyTrue It's not about the time of grounding - it's about not permanently grounding it, in which case the charge would go off as soon as you remove the hemisphere and try to measure it. $\endgroup$ – AV23 May 13 '15 at 10:00
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Your misconception seems to be that a conductor connected to the ground must have zero charge. The actually effect of grounding a conductor is to put it at zero potential, not charge. The ground is assumed to have an effectively unlimited supply of charges/capacity to accept charges in order to make this condition true.

After step 2 and before step 3, the electric field due to the charge on the inner sphere will cause a separation of charges in the outer sphere, inducing a charge density on the inner surface of the outer sphere of total charge equal to the negative of the charge on the inner sphere and (to ensure that at this point, the outer sphere is neutral) a charge density on the outer surface of the outer sphere of total charge equal to the the charge on the inner sphere. Then, in step 3, when you ground the outer sphere, the charge on the outer surface flows into the ground, but the charge on the inner surface is "bound" attractive the electric force due to the charge on the inner sphere, leaving a total charge on the outer sphere equal to the negative of the charge on the inner sphere at step 4.

Now, we can see quantitatively why the total charge left behind at step 4 is equal to the negative of the charge on the inner sphere. Suppose that the inner sphere has radius $R_1$, the outer sphere has radius $R_2$ and the total charge on the inner sphere is $Q$. By symmetry, the charge $Q$ is distributed uniformly on the inner sphere, and then by Coulomb's law and the shell theorem the potential due to the charge on the inner sphere at radius $r>=R_1$ is $kQ/r^2$. The dielectric material is neutral, so by a similar argument, it makes no contribution to the potential for $r>=R_2$. Thus the potential due to the inner sphere and the dielectric material at the surface of the outer sphere is constant at the value $kQ/R_2^2$. Now, the condition imposed by the grounding of the outer sphere is that the outer sphere be at zero potential, so there must be some induced charge density in the outer sphere to cancel out the potential due to the inner sphere. By symmetry considerations, the charge will once again be distributed in a spherically symmetric way on the inner surface of the outer sphere; let the total charge induced be $Q'$. Once again, by Coulomb's law and the shell theorem the potential due to the charge on the outer sphere at radius $r>=R_2$ is $kQ'/r^2$. Thus, by the principle of superposition the total potential at the outer surface of the outer sphere is $kQ/R_2^2+kQ'/R_2^2=k(Q+Q')/R_2^2=0$, since the outer sphere is grounded. Therefore, we must have $Q'=-Q$.

We could instead have worked more precisely in terms terms of the electric displacement field and considered the matching conditions at interfaces that relates the free surface charge density to the normal component of the electric displacement field, but the analysis would not change significantly. The root of your misunderstanding is thinking that a conductor connected to the ground must have zero charge (which is understandable, since your textbook uses the word "discharge"), instead of zero potential.

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