00:01
For this problem on the topic of kauser's law, we are told that an insulating sphere with radius a has uniform charge density.
00:08
The sphere is centered at r is equal to b, and we want to show the electric field inside the sphere is given by the expression provided.
00:15
We are then told that an insulating sphere of radius capital r has a spherical whole of radius little a located within its volume and centered at b from the center of the sphere.
00:27
Now, if this solid part of the sphere has a uniform volume charge density row, we want to find the magnitude and direction of the electric field inside the hole and show that this electric field is uniform over the entire hole.
00:40
So firstly, for part a, we know that for an insulating sphere of uniform charge density row and centered at the origin, the electric field inside the sphere e is given by the total charge q times r prime.
00:59
Over 4 pi epsilon 0 times capital r cubed where r prime here is the vector from the center of the sphere to the point where the electric field e is calculated now we also know that the uniform charge density row is equal to 3 times q over 4 pi r cubed so we can write the electric field strength e as simply row times r over 3 epsilon 0.
01:39
But we need to write this as a vector.
01:41
We know the electric field vector is readily outward in the direction of r prime.
01:47
So the electric field vector e is equal to row times vector r prime over 3 epsilon not.
01:58
Now for a sphere whose center is located by vector b, a point inside the sphere is located by and r is located by the vector r prime is equal to r minus b relative to the center of the sphere as shown in the following figure.
02:18
So we can clearly see that for a sphere centered at point b, the electric field e is equal to row into r minus b divided by three times epsilon not...