00:01
Hi there, so for this problem we need to use gauss law to find the electric field inside and upside spherical shield of radius capital r, which carries an uniform surface charge density, sigma, and we need to compare this answer to the problem 2 .7.
00:24
Now in this case, what we are going to do is to use two gaussian surface.
00:33
One that is inside the sphere and order that is upside the sphere.
00:41
So the first one that is going to be inside the sphere, something like this.
00:52
It's going to have a radius art, and then using the gaussian surface, we can conclude that inside the close -indexam.
01:07
The close integral of the product between the electric field and the differential in area is equal to the electric field.
01:19
It's magnitude because we can take out the electric field because it is a constant times the integral of the area, which is the total area that we know is 4 times pi times the radius square.
01:34
So we know that by gauss law, this is equal to the enclosed charge divided by epsilon sub -0.
01:45
And because we know that the enclosed charge is equal to zero, we can conclude that the electric field inside the sphere is equal to zero.
01:58
Now, upside the sphere, we choose a gaussian surface with a radius that is greater, than the radius of the sphere itself...