Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.8x10^4 N/C, as shown in Figure. Calculate the electric flux through (a) the vertical rectangular surface, (b) the slanted surface, and (c) the entire surface of the box.

Today we have a simple exercise about electric flux. Let's begin by recalling what the hell is the electric flux?.

Consider the definition that give us physics books to refresh our memories on the subject: Electric flux is proportional to the number of electric field lines that penetrate a surface.

If the electric field is uniform and makes an angle theta with the normal to a surface of area A, the electric flux through the surface is:


It is convenient (do this we allow us to work more comfortably and with less likelihood of errors) rename the magnitudes that gives us the picture of the problem, this to work on problem in algebraic symbolic form and not be dragging values throughout the problem resolution.
All right! we remember the definition of electric flux and rename the dimensions of the box we will now apply the definition of electric flux for solving the part a. Here they ask us the electric flux of rectangular vertical face. To apply the formula we should note that the angle between the electric field. and the normal vector to the surface is 180 degrees for this angle the cosine is -1. therefore the electric flux in this box lid is negative, let's see what is said in equations:

For part b. We do the same procedure as that used in the part a, we start by defining the angle between the normal vector to the surface and the electric field.

Then we find that the angle is 60 degrees. Now we need the length of the cap (which we called for practical purposes "w") to find "w" we use the definition of the cosine of an acute angle in a right triangle, we have the ingredients let us now the answer:

Now in part c, we need the electric flux present through all caps. See, the flux in the vertical rectangular cover was obtained in part a, the flux at the top of the box was obtained in part b.
We lack electric fluxes present in two triangular lids vertically to the sides of the box, it's electric flux is zero .. as simple as that.

Why?

Because there is no electric field lines penetrating these caps. Therefore the total electric flux over the entire surface of the box is zero, because the flux 1 and 2 are equal in magnitude but opposite in sign.
This was all in this post... if you like our work share it with your friends! see you later! and remember ... physics it's easy! very easy... ;)