Karl Friedrich Gauss is widely regarded as one of the greatest mathematicians of all time. Some consider him to be the greatest.Gauss discovered a way to think about Coulomb's law that allows us to calculate the electric field of charge distributions in a straightforward manner. Although Gauss did not discover a new physical law, his restatement of Coulomb's law was equally important. In this tutorial we will describe about electric flux, concept of eclectic flux, the flux of electric field, unit and dimensional formula of electric flux.
To comprehend Gauss' law, we must first grasp the concept of Electric Flux across a surface. It is a technique for calculating the total electric field that crosses a surface.
Electric flux, measured in coulombs, describes the force connected between positive and negative charges. An electric flux can be represented by flux lines emanating from positive charges and terminating on negative charges. A surface that emits electricity is measured by electric flux.
Surfaces are used to represent electric flux, along with values of electric fields at each points.To demonstrate how flux is determined, let’s begin with a simple case. An electric field constant (having the same vector value) is applied across a flat surface with area A.
The area vector A characterises the surface. This is a vector with magnitude A that points perpendicularly (normally) to the surface.
Actually, there is a minor issue here: As far as vector A is concerned, there are only two options. If this direction was chosen, it would be normal to the surface with the same magnitude regardless of whether it points in the opposite direction. However, in every problem involving electric flux, the direction chosen will be obvious.
The electric flux $\mathrm{\phi }$ in this simple case is given by
$$\mathrm{\phi =E.A=EAcos\theta }$$
It may be necessary to use advanced calculus methods to find the flux on a surface that isn't flat in general, and the electric field won't be uniform (in magnitude or direction) across it. However, the basic idea is straightforward. It is a good approximation to treat the surface as flat and to treat it as a continuous electric field if we divide it into small sections (for instance, squares) that are small enough to get a reasonable representation of the electric field. Let’s consider the $\mathrm{i^{th}}$ square is having area $\mathrm{\Delta A_{i}}$ and the electric field is near to $\mathrm{E_{i}}$. The electric flux for the small square is then
$$\mathrm{\Delta \phi_{i} =\Delta E_{i}.A_{i}}$$
It is approximately equal to adding up all the individual contributions to get the total electric flux for the entire surface.
$$\mathrm{\Delta \phi_{i} \approx\sum_{i}^{} \Delta E_{i}.A_{i}}$$
By increasing the number of little squares and reducing their size, we get close to the real value of electric flux$\mathrm{\phi}$. An integral (for one variable) can also be computed using a similar procedure in calculus. Here, we get a surface integral, and the electric flux over the surface S is
$$\mathrm{\phi =\oint E.dA}$$
Here's the simplest example: the field is constant, and the surface is perpendicular to it. By multiplying the surface area A by the electric field, one calculates the flux of electricity across the surface.
$$\mathrm{\phi = E.A}$$
The flux measures how many field lines cross the surface in this situation. It is expected that fewer lines would cross if the surface were tilted at an angle. Nothing crosses the surface if it is parallel to the lines.
The flux defined as crossing a surface whose normal is at an angle to the field lines is
$$\mathrm{\phi = EAcos\theta }$$
Every surface has a normal direction associated with it. The area of a surface can be thought of as a vector with a normal direction. The flux is calculated by taking the dot product of the electric field and this area vector.
$$\mathrm{\phi = E.A}$$
An electric field has a property called electric flux, which is the number of lines of force intersecting a given area. There is an assumption that electric field lines have positive charges at the beginning and negative charges at the end.
Those field lines pointing into closed surfaces are negative, while those pointing away from closed surfaces are positive. Whenever there is no net charge on a closed surface, field lines are directed through its interior and directed outward elsewhere. Electric fluxes are equal in magnitude when the net fluxes are zero, or when the magnitude of the negative flux equals the magnitude of the positive flux.
Electric flux, measured in coulombs, describes the force connected between positive and negative charges. An electric flux can be represented by flux lines emanating from positive charges and terminating on negative charges. A surface that emits electricity is measured by electric flux.
Unit of electric flux is $\mathrm{\frac{Newton\times (metres)^{2}}{coloumb}}$ or $\mathrm{\frac{Nm^{2}}{c}}$ and the dimension is $\mathrm{\left [ M^{1}L^{3}T^{-3} I^{-1}\right ]}$.
Electric flux is depend on the number of lines passing througha given area in an electric field. It is assumed that these lines are begin with positive charges and end with negative charges. Positive field lines point into closed surfaces, while negative field lines point away from closed surfaces. In this tutorial we have described about electric flux, concept of eclectic flux, the flux of electric field, unit and dimensional formula of electric flux.
Q1. Is electric flux affected by distance?
Ans. The flux is independent of the distance r. An enclosed surface surrounding a point charge would yield the same result, regardless of its size.
Q2. What do you mean by the flux of a wave?
Ans. It is the amount of energy that passes through a specific area normal to the direction of propagation of waves. It is denoted by S
Q3. Can there be a negative flux?
Ans. Remember that the number of electric field lines crossing a surface is proportional to the flux. Electric field lines will either enter or exit a closed surface, and thus the corresponding flux will be positive or negative.
Q4. What is the significance of electric flux?
Ans. The physical significance of electric flux is revealed in the Ampere-Maxwell law, which deals with displacement current. Electric flux is required to explain electric current continuity in capacitive circuits.
Q5. Is flux affected by velocity?
Ans. The flux is equal to the area projected perpendicular to the field and the magnitude of the field. In this case, the flux will change only if the plane's orientation relative to the direction of the field changes. As a result, the final answer is that the flux in this case is independent of velocity.
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