00:01
Hi there, so for this problem, we are told that two spherical cavities of rating a and b are hollowed out from the interior of a neutral conducting sphere of radius capital r, as is shown in this figure.
00:18
Now, at the center of each cavity, a point charge is placed, and we need to call this charge qa and qb.
00:27
Now, for part a of this problem, we need to find the surface.
00:31
Charges sigma a, sigma b, and sigma capital art for part a of this problem.
00:43
So in this case, we know that the surface charge density a is going to be minus the charge qa.
01:01
It is minus because we know that in this charge at the surface of a, it's going to be negative because the charge inside is positive.
01:11
So we're going to obtain 4 times pi times a to the square.
01:17
Now the surface charge density b is going to be minus again because it is in the opposite side of the charge that is given in this case, divided by 4 times pi times b square.
01:34
And finally, for the charge that we enclose, at the surface of the bigger sphere, the sphere of radius capital r, is going to be the sum of the two charges because we are enclosed in the two charges, divided by four times pi times the radius capital r to the square.
02:01
Now, that's a solution for part a of this problem.
02:06
Now for part b, we are asked about what is the field upside the conductor? door.
02:12
So to solve this, we use the fact that the electric field out is going to be 1 over 4 times pi times the sum of charges that we are enclosing.
02:27
So that is charge a plus the charge b and this divided by the separation distance art to the square.
02:37
And this in the radial direction where art is the vector from the center of a large sphere...