Charge q is placed at the corner of a cube. What is the flux through one face of a cube?
If charge is at the corner then it will be shared by 8 cubes.Charge for 1 cube = q/8Now in any given cube it is touching 3 9f its faces. So the area vector of that side and the electric field vector will be perpendicular. So flux through those 3 sides will be 0.Equal amount of flux will flow from the other 3 sides.So flux through 1 side = (q/8)/3 = q/24.Hope that helps.
What is the electric flux on each side of a cube if +q is placed on any one of the corners of the cube?
The key principle we first need to know is- Gauss’s law for electric flux. It states that the net electric flux through a closed surface is given by:where, Q is the net charge enclosed within the closed surface.Another key thing is, which is obvious, that electric flux, through surfaces which are symmetrical with respect to the geometry of the problem, will be equal. For example,if an electric charge , say +q, is placed at the center of the cube, then the net flux through all the faces should be equal to, as per Gauss’s law- [math]q/episilon0. [/math]And the net flux through each of the faces is- [math]q/(6episilon0)[/math], since the same flux should pass through each face as they are symmetrical to each other.Now, lets approach this problem.The first thing to observe is- faces 1,2 and 3 are symmetric with respect to the position of charge q. So equal amount of flux should pass through each of those faces.What about faces 4,5 and 6? Their planes contain the charge q. So no electric field is perpendicular to them, and hence the flux through faces 4,5 and 6, which is by definition,will be zero, as the angle between E and A is 90 degrees. Below image shows the scenario for a particular face(4,5 or 6).So, now we know that whole of the flux through the cube is passing through faces 1,2 and 3, and in equal amount. So, if F be the net flux through the cube, then through each face, net flux is: F/3.Now, lets find F. Now I am going to create symmetry around the charge q, as shown shown below.I have created 7 other cubes of same dimension as of original cube, the center of which is the position of the charge q. So by symmetry, net flux through through each of the 8 cubes should be the same, which is F.But the net flux through all the cubes will be, by Gauss’s law-[math]q/episilon0[/math]So, through each cube, net flux is: [math]q/8episilon0, [/math]which is equal to F.So, we arrive at the conclusion that, for the original problem geometry, where a charge q was placed at one corner of the cube, the net flux through each of the faces marked by 1,2 and 3, is F/3=q/24episilon0. And through faces marked by 4,5 and 6, the net flux is 0(zero).
Calculate the flux through just the top face of a box?
The top face of the box is on the surface z = 3 with normal <0, 0, 1>. Hence, the flux ∫∫s F · dS equals ∫∫
Calculate the flux through the shaded face of the cube?
Looking for any help regarding this physics problem. A cube of edge length = 9.0 cm is positioned as shown in the figure below. There is a uniform magnetic field throughout the region with components Bx = +9.0 T, By = +2.0 T, and Bz = +5.0 T. (a) Calculate the flux through the shaded face of the cube.
A charge Q is placed at the centre of a cube of side A. What is the electric flux through the cube?
According to Gauss’s law,[math]\displaystyle\oint_\,E \cdot ds= Q/ \epsilon[/math]Therefore the total flux through the cube is [math]Q/ \epsilon.[/math]The flux through each side of the cube is [math]Q/ 6\epsilon[/math].