When it comes to the study of science and the functioning of electricity, there is boundless knowledge and information that one stands to gain. The concept of Electric flux is one such field of study of science. It is pertinent to the understanding of electric force and its behavior. Let us study more about the concept of Electric flux.
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If there is no given netchargewithin a given closed surface then everyfield linedirected into the given surface continues through the interior and is usually directed outward elsewhere on the surface. The negative flux just equals in magnitude the positive flux, so that the net or total, electric flux is zero.
If a net charge is contained inside a closed surface, the total flux through the surface is proportional to the enclosed charge, positive if it is positive, negative if it is negative.
The mathematical relation between electric flux and the enclosed charge is known asGauss law for the electric field. It is one of the fundamental laws ofelectromagnetism. In the related meter-kilogram-second system and theInternational System of Units(SI) the net flux of an electric field through any closed surface is usually equal to the enclosed charge, in units ofcoulombs, divided by a constant, called thepermittivityof free space.
In the centimeter-gram-second system, the net flux of an electric field through any closed surface is equal to the consistent 4π times the enclosed charge, measured in electrostatic units (esu). Electric flux is proportional to the number of electric field lines going through a virtual surface. You can understand this with an equation.
If the electric field is uniform, the electric flux (ΦE) passing through a surface of vector area S is:
where E is the magnitude of the electric field (having units of V/m), S is the area of the surface, and θ is the angle between the electric field lines and the normal (perpendicular) to S. For a non-uniform electric field, usually the electric flux dΦE through a small surface area dS is denoted by:
wherethe electric field is E, multiplied by the component of area perpendicular to the field.
Question: An electric field of 500 V/m makes an angle of 30.00 with the surface vector. It has a magnitude of 0.500 m2. Find the electric flux that passes through the surface.
Solution: The electric flux which is passing through the surface is given by the equation as:
Notice that the unit of electric flux is a volt-time a meter.
Question:Consider a uniform electric field E=3×103îN/C. What is the flux of this field through a square of 10cmon a side whose plane is parallel to the yz plane?
