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A spherical shell, uniformly charged, exerts no field inside it ; the point being either at centre or off-centre.

According to Gauss law if we construct a spherical Gaussian surface all inside the shell, it contains no charge inside and thereby zero net flux on that surface. Now, here we usually conclude that the field is zero as flux is zero.

My question is how can we say the field is zero, whereas the flux can also be zero in other configurations, where ingoing and outgoing field lines are equal in number. For instance, spherically symmetrical $1/r^2$ field and the source point being outside Gaussian surface.

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One of the deep properties of space that physicists believe to be true is that space is isotropic; that is, in a vacuum one direction is no better than any other.

Isotropy holds equally well if the system in question is spherically symmetric: as long as there is no way to distinguish one direction from the other, you could not possibly expect the physics to behave in different ways in different directions...

This is very good news, because your uniformly charged shell system is spherically symmetric! The charged shell determines a natural origin at the center of the shell, but all directions from this origin are still equivalent, since a spherical shell has no structure that could determine a $``$special$"$ direction. Thus, at a given radius $r$ from the center of the shell, any physical quantity you may want to determine, including the electric field, must be the same in all directions.

(One way to formalize this argument is as follows: assume for contradiction that your uniformly charged shell results in an electric field that is not spherically symmetric. Then there must be some radius $r'$ for which $\vec{E}(r', \theta_1, \phi_1) \neq \vec{E}(r', \theta_2, \phi_2)$; that is, two different directions at the same radius give distinct results. But now rotate the charged shell in space so that $\theta_1 \to \theta_2$ and $\phi_1 \to \phi_2$. Since the charged shell is spherically symmetric, after this rotation nothing will have changed, meaning the electric field at each point in space should be the same as before. But this means that $\vec{E}(r', \theta_1, \phi_1) = \vec{E}(r', \theta_2, \phi_2)$, a contradiction! Thus, the electric field must depend only on $r$.)

With this spherical symmetry in hand, we can now apply Gauss's Law. Since the electric field must be the same for all directions, choosing our Gaussian surface to be a sphere with radius $R$ smaller than the radius of the charged shell, we have $$\unicode{x222F}E\cdot \hat{n} dS = 4\pi R^2E(R) = 0,$$ which of course implies that $E(R) = 0$ for all $R$ enclosed in the uniformly charged shell.

To recap, other field configurations are not possible because they would violate spherical symmetry and/or Gauss's Law. Using symmetries to solve problems in physics is a very powerful, but often quite subtle, skill, and one that is worth developing as soon as possible.

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  • $\begingroup$ Thank you for your careful and well-explained answer. $\endgroup$ – bitan maity Jul 18 '20 at 18:13
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TL;DR: We can say that the electric field is zero in the first case because of the symmetry of the charge configuration. This symmetry is not present in the second case you mention.

Think about it, in the case of a uniformly charged spherical shell of radius $R$, choose a spherical Gaussian surface inside. Suppose for a moment that there is a not null electric field in the region $r<R$. The modulus of $\vec{E}$ has be the same all over the spherical Gaussian surface (because of symmetry), and the direction would be either radially outward $\hat{r}$ or inward $-\hat{r}$ (again, because of symmetry). Then from Gauss' law, since there is no charge inside the Gaussian sphere, the electric flux through the sphere is zero

$$\int\vec{E}\cdot\vec{dS}=0.\tag{1}$$

Since $\vec{E}=E\hat{r}$ and $\vec{dS}=dS\hat{r}$ both have radial direction, $\vec{E}\cdot\vec{dS}=E\,dS$ and Gauss' law reduces to

$$\int E\,dS=0,\tag{2}$$ $$E\int dS=0,\tag{3}$$ $$E=0.\tag{4}$$

In the second case, however, if the source point is outside a general Gaussian surface, we can't say in general anything about the electric field in the surface, i.e., we can't say $|\vec{E}|$ has the same value at every point in the surface, neither can we say that the electric field is normal to the surface $\vec{E}\perp\vec{dS}$ at every point, so that, even if Gauss' law is still valid (Eq. $(1)$) we can't follow the steps in Eqs. $(2)$, $(3)$ and $(4)$.

So the bottom line is: in the first case the spherical Gaussian surface resembles the symmetry of the charge distribution; in the second case no. Take a look at my answer to this related question.

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