Electric field intensity due to a uniformly charged thin spherical shell

At a point outside the shell :

Let us consider a thin spherical shell of radius R on which positive Charge q is uniformly distributed [in thefigure given below].

A sphere of radius (r > R) is drawn with centre O as Gaussian surface. We are to find electric field intensity at P on this surface. According to Gauss’ theorem,

\oint_s \vec{E}.\vec{dS}= \frac{q}{\epsilon_0}

or, \oint_s EdS \cos\theta=\frac{q}{\epsilon_0}

As \vec and \vec {dS} are directed along the same direction, so \theta= 0°.

\therefore \int_sEdS=\frac{q}{\epsilon_0}

As E is same at all points on the Gaussian surface and directed radically outwards, so,

E\oint_s dS=\frac{q}{\epsilon_0} E\times 4\pi r^2=\frac{q}{\epsilon_0}

\therefore E=\frac1{4\pi\epsilon_0}\frac{q}{r^2} in SI and

E=\frac{q}{r^2} in CGS

This is same as the electric field due to a point charge q at a distance r, electric field due to a uniformly charged thin spiracle shell at a point outside the shell is such as if the whole charge were concentrated at the centre of the cell.

At a point on the surface of the shell :

When the point P is on the surface of the shell, i.e., when r = R, then we have

E=\frac1{4\pi\epsilon_0} in SI and

E=\frac{q}{r^2} in CGS

Electric field is maximum at the surface of the uniformly charged thin spherical shell.

At a point inside the shell :

When the point P is inside the charged spherical shell, i.e., r < R. So in this case charge resides outside the Gaussian surface and thus,

E=0

i.e., there is no electric field inside a uniformly charged spherical shell.

In the figure given below, it is shown the variation of electric field intensity (E) with the distance (r) from the centre of a uniformly charged spherical shell. It is seen that for r > R, E=0; for r = R, E=E_{max} and for r > R, E\propto \frac1{r^2} .