# Electrostatic Flux

**ELECTRIC FLUX**

** **The electric flux through a surface held inside an electric field represents the total number of electric lines of force crossing the surface in a direction normal to the surface.

Electric flux is a scalar quantity and it is denoted by $\varnothing $

In fact, flux is a property of a vector field and likewise, electric flux is associated with the electric field.

Relation between electric field intensity and electric flux

Where θ is the angle between the Electric field and the area vector

Thus, electric flux linked with a surface in an electric field may be defined as the surface integral of the electric field over that surface.

The unit of electric flux in Nm^{2} C^{-1}.

The electric flux through a surface is denoted by

${\oint}_{s}\overrightarrow{E.ds}$ = ${\oint}_{s}{E}_{n}dS$

**Gauss’ theorem**

It states that the total electric flux through a closed surface enclosing a charge is equal to $\frac{1}{{\xi}_{o}}$

Here, ${\xi}_{o}$ is the absolute permittivity of the free space.

If a closed surface S encloses an electric charge q, then according to Gauss’ theorem, the total electric flux through the closed surface is given by

$\varnothing =\frac{q}{{\xi}_{o}}$

By definition, the total electric flux through the closed surface S is given by Gauss’ theorem as

${\oint}_{s}\overrightarrow{E.ds}$ = $\varnothing =\frac{q}{{\xi}_{o}}$

**To prove Gauss’ theorem (For a spherically symmetric closed surface)**

** **Consider that a point electric charge q is situated at the centre of a sphere of radius r. Let

Where

Consider a small area element $\overrightarrow{dS}$

Therefore, electric flux through the closed surface of the sphere,

Now,

$\varnothing =\frac{1}{4\mathrm{\pi}{\xi}_{O}}.\frac{q}{{r}^{2}}\times 4{\mathrm{\pi r}}^{2}=\frac{\mathrm{q}}{{\mathrm{\xi}}_{\mathrm{o}}}$

**Gaussian surface**

The Gaussian surface around a charge distribution (may be a point charge, a line charge, a surface charge or a volume charge) is a closed surface, such that electric field intensity at all the points on the surface is same and the electric flux through the surface is along the normal to the surface.

Or total flux linked with a surface is $\frac{1}{{\epsilon}_{0}}$times the charge enclosed by the closed surface.

The electric field can also be calculated by Coulomb’s law, but Gauss’s law is easier. Besides, Gauss’s law is just a replica of Coulomb’s law. If we apply Gauss’s theorem to a point charge enclosed by a sphere, we will arrive at Coulomb’s law.

## How to Find Electric Field Using Gauss’s Law?

These include a few steps.

1. Initially, we should find the spatial symmetry (spherical, cylindrical, planar) of charge distribution.

2. Next, we need to find the Gaussian symmetry, same as that of the symmetry of spatial arrangement.

3. Find the integral along the Gaussian surface and then find the flux.

4. Find charge enclosed by Gaussian surface.

5. Find the electric field of charge distribution, the electric field due to a point charge.

The electric field is a vector field which is associated with the Coulomb force experienced by a test charge at each point in the space to the source charge. The magnitude and the direction of the electric field can be determined by the Coulomb force F on the test charge q. If the field is created by a positive charge, the electric field will be in a radially outward direction, and if the field is created by a negative charge, the electric field will be in a radially inwards direction.

Let a point charge Q be placed in a vacuum. Then, if we introduce another point charge q (test charge) at a distance r from the charge Q,

Then, the electric field at point p due to the point charge Q is given by,

The direction of the electric field due to a point charge Q is shown in the above figure. The magnitude of the electric field is proportional to the length of E. If a test charge, which is relatively larger, is brought within the area of the source charge Q, it is bound to modify the original electric field due to the source charge. A simple way to escape from this conflict is to use a very negligible test charge q.

Then, our definition of electric field modifies to,

According to this definition,

The electric field at point P due to point charge Q is,

## Electric Field Due to Line Charge

One application of Gauss’s law is to find the electric field due to the charged particle. Electric field due to line charge can be found easily by using Gauss’s law. Consider,

A line charge is in the form of a thin charged rod with linear charge density λ.

To find the electric intensity at point P at a perpendicular distance r from the rod, let us consider a right circular closed cylinder of radius r and length l with an infinitely long line of charge as its axis.

The magnitude of the electric field intensity at every point on the curved surface of the Gaussian surface (cylinder) is the same since all points are at the same distance from the line charge.

Therefore, the contribution of the curved surface of the cylinder towards electric flux,

2 Π rl is the curved surface area of the cylinder.

On the ends of the cylinder, the angle between the electric field and its direction is 90^{o}. So, these ends of the cylinder will not have any effect on the electric flux.

Therefore,

ϕ E =$\frac{q}{{\xi}_{0}}$

Charge enclosed in cylinder=line charge density × length= λ l, so according to Gauss’s law,

ϕ E =$\frac{q}{{\xi}_{0}}$

E(2 Π rl)= $\frac{\lambda l}{{\xi}_{0}}$

Or,

E=$\frac{\lambda}{{h}_{o}(2\Pi {\epsilon}_{0}r)}$

## Electric Field Due to Ring

We want to find the electric field at an axial point P due to a uniformly charged ring, as shown in the figure:

The centre of the ring is at point O.

The circumference of the circle makes an angle θ with line OP drawn from the centre of the ring to point P.

Now, consider a small element from the ring as dq, and calculate the field at point P due to this charge element.

We know that the electric field at point P is,

## Electric Field Due to Continuous Charge Distribution

Here, we need to consider that the charges are distributed continuously over a length or a surface or a volume.

If we want to find the electric field with a surface which carries charges continuously over the surface, it is not possible to find the electric field due to each charged constituent. So, in order to solve this problem, we consider an elementary area and integrate it.

If the total charge carried by an area element is equal to Δ Q, then the charge density of the element is,

*v *whose charge density is ρ. Let the distance of the volume element from point P is r. Then, the charge in the volume element is ρ Δv. Thus, the electric field is,

## Electric Field Due to a Uniformly Charged Sphere

Let σ be the uniform surface charge density of the sphere of radius R.

Let us find out the electric field intensity at a point P outside or inside the shell.

**Field outside the Shell**

We have to find the electric field intensity at a point P outside the spherical shell such that, OP = r.

Here, we take the Gaussian surface as a sphere of radius r.

Then, the electric field intensity is the same at every point of the Gaussian surface directed radially outwards,

So, according to Gauss’s theorem

So, it is clear that electric intensity at any point outside the spherical shell is such as if the entire charge is concentrated at the centre of the shell.

**Field at the Surface of the Shell**

Here, we have, r = R

Therefore,

E = $\frac{q}{(4\pi {R}^{2}{\epsilon}_{0})}$

If σ C/m^{2} is the charge density on the shell,

Then,

q=4 Π R^{2}. σ

Therefore,

E = (4 Π R^{2}. σ ) / (4 Π R^{2}. ε_{O}) = σ /ε_{0}

**Field Inside the Shell**

If point P lies inside the spherical shell, then the Gaussian surface is a surface of a sphere of radius r.

As there is no charge inside the spherical shell, the Gaussian surface encloses no charge. That is q = 0.

Therefore,

E = 0

Hence, the field inside the spherical shell is always zero.