 # Gaussian Surface

## What Is Gaussian Surface?

An arbitrarily closed surface in three-dimensional space through which the flux of vector fields is determined is referred to as the Gaussian surface. Their vector field referred here could either be a magnetic field, gravitational field or electric field. Below examples mostly considered an electric field as a vector field. Gauss Law calculates the gaussian surface.

$\Phi_{E}$   =        $\mathop{{\int\!\!\!\!\!\int}\mkern-14mu \bigcirc}$ ${_{\partial V}E.dA = \frac{{Q\left( V \right)}}{{{ \in _0}}}}$

Above formula is used to calculate the Gaussian surface. Q(V) refers to the electric charge limited in V.

Let us understand Gauss Law. In real terms, Gauss meaning is a unit of magnetic induction equal to one-tenth of tesla. In physics, Gauss Law also called as Gauss’s flux theorem. This law relates the distribution of electric carrier, i.e., charges following into electric field.  Like mentioned earlier, the surface considered may be closed, confining the volume such as spherical or cylindrical surface. Gauss’s Law is the combination of divergence theorem and Coulomb’s theorem.

To give you an idea Coulomb’s theorem is the charges opposite in nature attracts. In contrast, the same charges repel with  force, which is directly proportional to the product of charges and inversely proportional to the square of the distance between the charges. Divergence theorem is also known as Gauss theorem. It is an alternate way to defining the behaviour of the flux or flow of some physical quantity. If this quantity is conserved in some manner, the inverse square law is of the same notion as Gauss Law mentions conservation of flux directly.

With this understanding, let us, deep-dive, determine the Gaussian surface of various closed surfaces like sphere or cylinder.

### The Gaussian Surface of A Sphere

There could be several reasons when flux or electric field generated on the surface of spherical Gaussian surface –

• A single point charge

• Uniform distribution on a spherical shell

• Distribution of charge with spherical proportion or symmetry

Imagine a below sphere of radius R, where Q is the charge uniformly distributed. The electric field at a distance ‘r’ would be determined using Gauss Law.

$\Phi_{E}$ = $\mathop{{\int\!\!\!\!\!\int}\mkern-14mu \bigcirc}_{\partial s}$ = $\mathop{{\int\!\!\!\!\!\int}\mkern-14mu \bigcirc}_{c}$  EdA $Cos^{o}$ = E $\mathop{{\int\!\!\!\!\!\int}\mkern-14mu \bigcirc}_{s}$ dA

The surface area of a sphere is  $\iint {_sdA = 4\pi {r^2}}$, this implies$\Phi E = E4\pi {r^2}$. By Gauss Law , flux is also $\Phi E = \frac{{QA}}{{{ \in _o}}}$. Equate the above two expression simplifies the equation

$\Phi E\pi {r^2} = \frac{{QA}}{{{ \in _o}}} \Rightarrow \Phi E = \frac{{QA}}{{4\pi {r^2}{ \in _o}}}$

### The Gaussian Surface of The Cylinder

A closed cylindrical surface considered to determine vector field or the flux generated in the due to below parameters-

• Uniform charge present on the uniform infinite long line

• Uniform charge on the infinite plate

• Uniform charge on an infinitely long cylinder

Consider a point charge P present at a distance r containing charge density λ of an unlimited line charge. The rotational axis for the cylinder of length ‘h’ is the line charge, following is the charge q present inside in the cylinder:

$q = \lambda h$

Subsequent is the flux out of the cylindrical surface with the differential vector area dA on three different surfaces a, b and c are given as:

The Gaussian surface of the cylinder

$\Phi_{E}$ = $\mathop{{\int\!\!\!\!\!\int}\mkern-14mu \bigcirc}_{A}$ E. dA =  $\mathop{{\int\!\!\!\!\!\int}\mkern-14mu \bigcirc}_{a}$ E. dA + $\mathop{{\int\!\!\!\!\!\int}\mkern-14mu \bigcirc}_{b}$ E. dA + $\mathop{{\int\!\!\!\!\!\int}\mkern-14mu \bigcirc}_{c}$ E. dA

The Equation for Gaussian surface of the cylinder

ΦE=∬aEdAcos900+∬bEdAcos90o+∬cEdAcos0o=Ec dA c dA = 2πrh (which is the surface area of the cylinder)

(which is the surface area of the cylinder)

$\Phi E = E2\pi rh$$\Phi E = \frac{q}{{{ \in _o}}}$ (by Gauss law)

$E2\pi rh = \frac{{\lambda h}}{{{ \in _o}}} \Rightarrow E = \lambda 2\pi { \in _o}r$

The above equation showcases the cylindrical Gaussian surface with a distribution of charges uniformly.

### What Is A Gaussian Pillbox?

This surface is commonly aiding to find the electric field due to an infinite sheet of charge with uniform charge density, or a slab of charge with some finite thickness. Imagine a box in cylindrical shape comprising three components: the disk at one end of the cylinder with area πR², the disk at the other end with equal area, and the side of the cylinder. According to Gauss Law, the addition of the electric flux via the above component of the surface is proportionate to the enclosed charge of the pillbox. The field close to the sheet can be estimated constant; the pillbox tilted in such a manner where field lines pierce the disks at the ends of the field at a right angle, and the side of the cylinder is parallel to the field lines.

### Fun Facts

• Do you know the surface which is invalid Gaussian surface? Well, surfaces like a disk, hemisphere, the square cannot be Gaussian surfaces as these surfaces do not include three- dimensional volume and have boundaries. Infinite planes can be an approximate Gaussian surface.

• After reading the Gaussian surface of sphere and cylinder, you must be thinking of limitations of Gauss law. Yes, you are right!. Gauss Law has a specific restriction.Gauss law is limited to calculate the vector field of a closed surface. This law cannot determine the field due to electrical dipole. Thinking what an electrical dipole is? Well, the electrical dipole is nothing but a separation of positive and negative charge. For example, an electret is a permanent electric dipole.