00:01
Hi there, so for this problem, we are told to use gauss law to find the electric field inside a uniformly charged sphere with charge density row.
00:22
And we need to compare this with your answer of problem 2 .8.
00:28
Now, in this case, we're going to have the following.
00:32
We have this sphere and then what we are going to do is to choose a gaussian surface that is inside the charge sphere.
00:49
Now with that we can use gauss law to write that the close integral of the product between the electric field and the differential in area is equal to so because the electric field is a constant we can take it out of the integral and then we will have the integral of the different area that we know it is the area of the total of the sediment that we have for the sphere so we will have the magnitude of the electric field times the area which is four times pi times the radius square and this is equal by gauss law as the enclose charge divided by epsilon sub zero.
01:49
So with this we can write we know that the charge density is equal to the total charge divided by the volume.
01:59
Now since we want the enclosed charge we will have that the charge enclosed charge is equal to the volume.
02:10
We know that the volume of a sphere is four divided by three times pi...