00:01
For this problem on the topic of electric fields, we are told to consider a charge distribution as shown in the figure.
00:08
We want to show firstly that the magnitude of the electric field at the center of any face of the cube has the value given.
00:16
2 .18 k .e times q over s squared.
00:19
Secondly, we want to know the direction of the electric field at the center of the face, of the top face of the cube.
00:27
So drawing our cube, we can also see that there are zero contributions from the same face due to symmetry, but the opposite faces contribute 4 into k -e times q over r squared times sign of the angle phi.
00:51
And here we know that r with some geometry is equal to the square root of s over 2 alt squared plus s over 2 alt squared plus s squared and so we get r to be 1 .22 times s.
01:21
So we can express r in terms of s.
01:24
We also know that sine phi is equal to s over r...