00:01
We are looking to find the charge enclosed in the cube with x, y, z vertices, all of plus and minus one.
00:08
So we know that the net charge is given by the following formula.
00:16
And s has to be divided into six parts to represent the six faces of the cube.
00:22
So we have to find the flux of x equals 1, negative 1, y1, y, 1, y, negative 1, and z 1, z, negative 1.
00:34
Starting with x equals 1, this will give that n is equal to, remember that we, ds can be rewritten as nds, and n in this case must be equal to i because we're looking at when x is 1i.
00:48
So we're going to take the function multiplied by i.
00:52
So we're only going to be integrating the i term, integral of s of ds.
01:02
And that just becomes from when you dot two vectors together, one vector has no j or k terms.
01:08
The j and k terms of the other vector will go away.
01:12
And since x is going to be one for all points, we can rewrite this as the surface integral of just ds, which is going to be equal to the integral of the face of the cube.
01:24
And since x is plus and minus one, the integral of the face of that cube is going to have both the base and height of two...