Question

(i) Using Gauss Theorem shows mathematically that for any point outside the shell, the field due to a uniformly charged spherical shell is the same as the entire charge on the shell, is concentrated at the centre.
(ii) Why do you expect the electric field inside the shell to be zero according to this theorem?

Answer 1

Hint:Use Gauss’s law to find the electric field outside the spherical shell and compare the expression for a point charge. Gauss’s law states that the total flux due to a volume is proportional to the charge enclosed.

Formula used:
Gauss’s law is given by,
\[\phi = \oint\limits_s {\vec E} .d\vec S = \dfrac{q}{{{\varepsilon _0}}}\]
where, \[\vec E\] is the electric field, \[S\] is the closed surface placed in the electric field.
Total charge of continuous charge distribution,
\[q = \int {\sigma ds} \]
where, \[\sigma \] is the volume charge density and \[ds\] is the volume element.

Complete step by step answer:
We know that the electric field at a point due to charge is nothing but the force acting on a unit positive charge due to the charge at that point. For multiple charges since the electric field follows the superposition principle it is the sum of the electric field due to each charge. For a continuous charge distribution it is the integration over the source points. Here we have a shell which is uniformly charged. So, let the charge density per unit area be \[\sigma \] and the radius of the shell is R.

(i) Now, if we consider a Gaussian surface at a distance r from the center of the shell, then according to Gauss’s law we can write,
\[\phi = \oint\limits_s {\vec E} .d\vec S = \dfrac{{\int {\sigma ds} }}{{{\varepsilon _0}}}\]
Now, the electric field due to the shell must be uniform since the charge distribution is uniform and symmetric. So, we can write,
\[E\oint\limits_s {dS} = \dfrac{{\sigma \int {ds} }}{{{\varepsilon _0}}}\]
Now, the \[\oint\limits_s {dS} \]is at a radius r hence the surface area will be, \[4\pi {r^2}\]
And the \[\int {ds} \] is over the source points at a distance over R so the area will be, \[4\pi {R^2}\]
Hence, we will have,
\[E4\pi {r^2} = \dfrac{{\sigma 4\pi {R^2}}}{{{\varepsilon _0}}}\]
\[\Rightarrow E = \dfrac{{\sigma \times 4\pi {R^2}}}{{4\pi {r^2}{\varepsilon _0}}}\]
Now, the value of \[\sigma \times 4\pi {R^2}\] is equal to the total charge on the spherical shell. Hence, if we say the charge on spherical shell is Q then we can write,
\[\therefore E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}}\]
Also, we know that the electric field due to a point charge $Q$ is \[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}}\]. So, we can say for a spherical shell the electric field outside is the same as the entire charge on the shell if concentrated at the centre.

(ii) Now for a point inside the shell if we consider a Gaussian surface at a radius $r$ and \[r < R\], then the charge inside the spherical surface will be zero. Since all the charges are at radius $R$. Now, According to Gauss’s law we will have,
\[E\oint\limits_s {dS} = 0\]
Hence, the electric field will be zero since the surface integration cannot be zero.

Note: Remember that the total flux due to any volume is equal to the charged enclosed, the charges outside do not contribute the flux for a volume. For charges outside the volume the flux is always equal to zero. When calculating the integration always keep in mind the limits of the integration or the surface element is of which surface. Changing the limit or confusing the surface with any other surface, will give an incorrect solution.