Question

Write down Gauss's theorem of electrostatics. Find out intensity of the electric field at a point outside a uniformly charged thin spherical shell with its help.

Answer 1

Whenever there is a charged point located anywhere in space, the number of tubes of induction is equal to the magnitude of the charge.
If we bring a small positive test charge near the point charge, then the force acting on it is shown by lines known as lines of force. A bunch of lines of force is known as tube of induction.
Whenever there is a charged point located anywhere in space, the number of tubes of induction is equal to the magnitude of the charge.
Consider a point charge surrounded by a 3 dimentional closed surface. Thus the charge is inside the shell.
The total number of tubes of induction passing normally through that area is known as Total Normal Electric Induction (TNEI).
Gauss's theorem states that TNEI around any closed surface is equal to the sum of the total charges enclosed by it.
Mathematically, $TNEI=\Sigma q$
We know that $TNEI=\int \epsilon Ecos\theta ds$
Here $\theta$ is the angle between the area vector (perpendicular to the area) of area ds and ds is a very very small piece of the closed surface area.
Thus from all above information, we have :-
$\int \epsilon Ecos\theta ds=\Sigma q=Q$
In order to find the intensity of the electric field at a point outside a uniformly charged thin spherical shell, we need to imagine a Gaussian spherical surface, inside which the given sphere is present.

Here we assume that the sphere is conducting and hence the charges are uniformly spread on it.

We already have $Q=\int \epsilon Ecos\theta ds$

Due to symmetry and uniformity, $\theta$ is always = $0$, and $\therefore cos\theta =1$ always

$Q=\epsilon E\int ds=\epsilon E\times 4\pi r^2$ Since, area of sphere =$4\pi r^2$

$\therefore E=\frac{Q}{4\pi \epsilon r^2}$

This is the final formula.

We can notice that a charged conducting sphere behaves like a point charge.