Electric Field due to a Uniformly Charged Thin Spherical Shell ( Hollow Sphere)

Question: Using Gauss's theorem , derive expression for electric field intensity due to a Uniformly Charged hollow sphere (a) at a point outside the shell (b) at a point on the surface of the shell (c) at a point inside the shell

Solution:

(a) Electric field intensity at a point outside the shell : 

Consider a positive charge q distributed uniformly on the surface of a spherical shell of Radius R.

Let P is a point outside the shell at a distance r from the centre of the shell ( r>R) , where electric field is to be calculated.

According to Gauss's theorem 

    $\oint_{S} \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_{0}}$

   $ \oint_{S} E \, dS \cos \theta = \frac{q}{\epsilon_{0}}$ ....(i)

Since $\vec{E}$ and $d\vec{S}$ are along the same direction, so $\theta = 0^\circ$.

 $\therefore \oint_{S} E \, dS \cos 0^\circ = \frac{q}{\epsilon_{0}}$

or  $\oint_{S} E \, dS = \frac{q}{\epsilon_{0}}$

Since E at all points on the Gaussian surface is same and directed radially outwards, 

therefore,$E \oint_{S} dS = \frac{q}{\epsilon_{0}}$.  ...(ii)

But$\oint_{S} dS$= surface area of the spherical Gaussian surface 

$\oint_{S} dS = 4\pi r^2$

    $E \times 4\pi r^2 = \frac{q}{\epsilon_{0}}$

    $E= \frac{1}{4\pi \epsilon_{0}} \frac{q}{r^2}$.     ....(iii)

which is the electric field due to a charged spherical shell at a distance r.

In vector form, 

$\vec{E}= \frac{q}{4\pi \epsilon_{0} r^2} \hat{r}$ ....(iv)

Thus, electric field due to a uniformly charged thin spherical shell at a point outside the shell is such as if the whole charge were concentrated at the centre of the shell.

(b) Electric field intensity at a point on the surface of the shell

When point of observation (P) is at the surface of the shell then $r = R$.

Therefore, eqn. (iii) becomes

    $E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{R^2}$. ....(v)

Now ,$q = \sigma \times 4\pi R^2$. Where $\sigma$ = surface charge density of the charge on the shell

 $E = \frac{1}{4\pi \epsilon_{0}} \frac{\sigma \times 4\pi R^2}{R^2}$

$E = \frac{\sigma}{\epsilon_{0}}$ .....(vi)

(c) Electric field intensities at a point inside the shell

Let P be the point inside the shell at a distance r' (r' < R) from the centre of the shell. Draw a Gaussian sphere of radius r' with O as the centre.

According to Gauss' theorem,

   $\oint_{S} \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_{0}}$

Here, Gaussian surface encloses no charge i.e., q = 0

   $ \oint_{S} \vec{E} \cdot d\vec{S}= 0$ 

or E = 0 ....(vii)

Thus, there is no electric field inside a uniformly charged thin spherical shell.

Variation of Electric field intensities (E) with distance (r) from the centre (O) of the charged spherical shell is

[(i)] When, r < R, then E = 0

[(ii)] When, r = R, we get, 

$E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{R^2} = \frac{\sigma}{\epsilon_{0}}$ (i.e., maximum value)

[(iii)] When, r > R, we get, 

$E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{r^2}$ 

or $E \propto \frac{1}{r^2}$

Thus, electric field intensity is zero inside the charged spherical shell. It is maximum at the surface of the charged spherical shell and then varies inversely with (distance)$^2$ away from the surface of the charged shell.