Electric flux formula cube1/23/2024 ![]() ![]() Here you will get to know in detail about:Įlectric flux through an elementary area, ds is defined as the scalar product of area and field, i.e. We will study its application in detail here. We will also discuss Gauss’s Law in detail which is an application of Electric Flux which helps to calculate Electric Field for a given charge distribution enclosed by a closed surface. It is rate of flow of electric field through a surface which can be open or closed. unit of electric flux is volt metres (V m) and the dimensions of the electric flux are - Kg m 3 s-3 A-1 or NC-1 m 2. ![]() The magnitude of the flux through rectangle BCKF is equal to the magnitudes of the flux through both the top and bottom faces. ![]() Here, the net flux through the cube is equal to zero. The direction of the vector of area elements, is perpendicular to the surface itself. The net electric flux through the cube is the sum of fluxes through the six faces. Also we will further discuss the Gauss-theorem and apply the formula of it to find out the desired answer. The orientation of the area elements impacts the amount of electric flux, hence the area element is a vector quantity. It is a property of Electric Field which tells us the number of field lines crossing a particular area. Hint : In order to answer this question, to know about the electric flux associated with one of the faces of the cube we will first know about the electric flux. Or there is no charge enclosed within the closed surface. Since electric field is uniform, it is created by a source very far from the closed surface. According to Gauss's Law, flux through a closed surface is given by: E.dS oq encosed. For Class 10 th Boards + JEE/NEET Studentįor Class 9 th + 10 th + JEE/NEET StudentĮlectric lines of Force are used to measure Electric Flux. Electric flux through a closed surface in uniform electric field. ![]()
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