00:01
Hi there, so for this problem, the solid metal sphere of radius a equals to 0 .2 meters shown in the figure.
00:17
This figure right here has a surface charge distribution sigma, and the potential difference between the surface of the sphere and a point p at a distance are p equals to 0 .5 meters as is shown in the figure from the center of the sphere is equal to the potential at the surface minus the potential at the point p is equal to four pi times a four pi balls which is equal or approximate to 12 12 .56 balls.
01:16
So we need to determine the value of the surface charge distribution sigma.
01:25
So the potential of a sphere is given as we know by the following equation.
01:34
That is the charge divided by 4 times pi times epsilon times the radius are, where q is a total charge of the sphere.
01:51
The total charge is given by 4 pi times the radius of the surface to the square times the charge distribution sigma, where rs is the radius of the sphere.
02:13
So the potential difference between the surface of the surface of the sphere.
02:17
And the point p is equal to the following.
02:28
We will have q over 4 times pi times epsilon sub 0.
02:35
As you can see, we are taking out everything that is common, and we will have 1 over the radius rs minus 1 over the radius rp.
02:47
And we just simply substitute what is the definition of the charge q in terms of the charge distribution...