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enter image description here

this is a screen shot of a past paper exam, sorry that I could not get better quality.

I am struggling to answer this question (b) and (c), I believe that once I solved for (c) it should be easy to get (d)

If this was the other way round(metal ball surrounded by a rubber dielectric) the I could solve it but this way seems tricky.

I would really appreciate it if you could help me out

I believe that for a metal(conductor) to have a Q charge total spread over there must -Q on the underside of metal shell(due to negligible thickness and no other charge, but possibly what the metal shell could be covering), thus there must be an overall Q source charge inside the rubber because the sum of surface bound charge and volume charge should = 0. But this doesn't help my case it seems

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Using the Gauss's Law given in (a) and the spherical symmetry of the problem, you can calculate $\vec{D}$ everywhere. Note that this calculation is possible with the knowledge of only $\textit{free}$ charges. The only free charge present in this problem is the charge $Q$ spread uniformly over the metal's surface. The remaining steps are straightforward.

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  • $\begingroup$ is D zero inside the metal shell? $\endgroup$
    – Reuben
    Sep 15 at 20:00
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    $\begingroup$ The metal is assumed to have zero thickness, so we can't technically give the value of $D$ $\textit{inside}$ the metal shell. Right below the metal's surface, $D$ will be zero. $\endgroup$
    – Quantour
    Sep 15 at 20:02
  • $\begingroup$ thank you, but I am slightly confused, using E=(permitivity)*D, then E = 0 below metal? $\endgroup$
    – Reuben
    Sep 15 at 20:07
  • $\begingroup$ Or is E= -P/(permittivity of free space) $\endgroup$
    – Reuben
    Sep 15 at 20:10
  • $\begingroup$ $D=0$ is equivalent to $E=-\frac{P}{\epsilon_0}$. In case of the linear dielectric, $E=P=0$ holds. $\endgroup$
    – Quantour
    Sep 15 at 20:20

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