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Suppose we have a spherical insulator charged with $Q$ inside a spherical conducting shell charged with $-2Q$.

My textbook says that:

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But I cant figure out why is that.

All I know that as a conductor in electrostatic equilibrium the field inside it is zero. So if we take a concentric gaussian surface with a radius $b<r<c$ we see that the total charge inside it has to be zero. And since the charge lies on the surface of the shell the inner walls have to be charged with $-Q$.

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    $\begingroup$ The field inside the metal of a conductor is zero, not necessarily inside a cavity in the conductor $\endgroup$ – Lelouch Sep 2 '16 at 14:52
  • $\begingroup$ I know but the textbook says that in the cavity the field that the conductor creates is zero right? $\endgroup$ – George Smyridis Sep 2 '16 at 15:21
  • $\begingroup$ The field that the conductor creates is zero, but there is a field due to the sphere carrying charge Q. $\endgroup$ – GeeJay Sep 2 '16 at 16:06
  • $\begingroup$ Yeah I TOTALLY agree. What i am asking is WHY is the field that the conductor creates is zero? I know that when there is no sphere inside that the field is zero. But when there is why is it zero? $\endgroup$ – George Smyridis Sep 2 '16 at 17:05
  • $\begingroup$ The textbook that is being referred to here is "Physics for Scientists and Engineers with Modern Physics-Raymond A. Serway, John W. Jewett-Cengage Learning (2013)" $\endgroup$ – claws Sep 15 '17 at 17:06
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We can see why the conductor doesn't contribute to the electric field inside a cavity by using the fact that the electric field is conservative in electrostatics. $\vec E$ being conservative implies that $\oint_C d\vec s\cdot \vec E = 0$ for any closed curve $C.$

Constructing a Gaussian surface $S$ around the cavity yields $\iint_S d\vec A\cdot \vec E = 0$, but that doesn't mean there aren't equal amounts of positive and negative charges on the surface of the cavity, necessarily. We can only conclude that the net charge in the cavity is zero.

Imagine there was some nonzero field inside a cavity, beginning on some positive charge and ending on the negative charges. Then, pick a closed path $\Gamma$ that passes through the cavity with portion $\gamma_1$, but also passes through the "meat" of the conductor with portion $\gamma_2$ and joins up on itself such that $\gamma_1 + \gamma_2 = \Gamma$. The contribution to the integral is zero over $\gamma_2$ since the field in the "meat" of a conductor is zero. It follows that the line integral of the field $\vec E$ over $\Gamma$ is $$ \oint_\Gamma d\vec s\cdot \vec E = \int_{\gamma_1}d\vec s\cdot \vec E + \int_{\gamma_2}d\vec s\cdot \vec E = \int_{\gamma_1}d\vec s\cdot \vec E \ne 0, $$ since the field $\vec E$ is nonzero over $\gamma_1$. But this is clearly a contradiction of the fact that $\oint_C d\vec s\cdot \vec E = 0$ in electrostatics. We are forced to conclude that if there is any field in the cavity of a conductor that it is due to the presence of other bodies.

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All I know that as a conductor in electrostatic equilibrium the field inside it is zero. So if we take a concentric gaussian surface with a radius $b<r<c$ we see that the total charge inside it has to be zero. And since the charge lies on the surface of the shell the inner walls have to be charged with $−Q$.

That is all correct.

Apply Gaussian surfaces in regions 2, 3 and 4. The charged enclosed by surface 2 is $+Q$ so there is a field $E(r)=kQ/r^2$ in the region $a<r<b$. A charge $-Q$ is induced on the inner surface of the conducting shell, so the total charge enclosed by surface 3 is $0$. Hence the electric field in region 3 $(b<r<c)$ - ie inside the conducting shell - is $E(r)=0$.

The total charge on the conducting shell is $-2Q$. There is already a charge $-Q$ on the inner surface so the charge on the outer surface is $-Q$. Surface 4 encloses net charge $-Q$ hence the field in region 4 $(r>c)$ is $E(r)=-kQ/r^2$.

If there were no charged sphere 1 inside the spherical shell, the charge enclosed by both surfaces 2 and 3 would be $0$, so there would be no field in regions 2 or 3 - ie $E(r)=0$ for $0<r<c$.

In both cases the field inside the conductor itself is $0$, regardless of whatever charges are inside or outside the conductor, or on the inner or outer surfaces. The field inside a cavity enclosed by the conductor is only zero if there is no charge in the cavity.


The charge on the conducting shell creates zero electric field in the region $r<b$, so the conducting shell has no effect on the field in region 2 due to the sphere.

Your textbook is applying the Superposition Principle : the field in the various regions is the superposition of the separate fields from the charges when considered separately.

The field in region 2 due to the charge on the shell is zero, but the field in region 2 due to the charge on the sphere is $E(r)=+kQ/r^2$. The total field in this region is the sum of these two fields. The same argument applies for region 4, where the net electric field is $E(r)=+kQ/r^2-2kQ/r^2=-kQ/r^2$.

If we assume that the entire charge of $-2Q$ on the shell remains on its outer surface - which is where it would be if the inner charge did not exist - then by the Superposition Principle the field inside the shell $(b<r<c)$ should be $E(r)=+kQ/r^2$ - which contradicts the principle that the electric field inside a conductor is zero.

The difficulty of applying the Superposition Principle to region 3 (inside the shell) is that the principle assumes that charges do not move when other charges are brought into proximity to them. This is true for the inner sphere - the charge remains on its outer surface when the charged conducting shell is brought to $r=c$ from $r=\infty$. But it is not true for the shell.

If the shell had infinitesimal thickness, there would be no difference between the inner and outer surfaces, so in effect this charge cannot move. But if the shell has finite thickness, charge can move from one surface to the other.

Whether the charge on the shell is on the outer or inner surface makes no difference to the electric field it generates in the cavity inside the shell $(r<b)$, but it does have an effect on the electric field inside the shell itself $(b<r<c)$. The electric field in this region is now the sum of that from the sphere $E(r)=+kQ/r^2$ and that from the charge on the inner surface of the shell $E(r)=-kQ/r^2$. That sum is zero.

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