00:03
Given the illustration here, we are asked to find the electric potential at a point, let's say, x, at the lower right corner of the system.
00:17
Now, we are given the distances between the charges, the magnitude of the charges, where q is equal to negative 2 .0 -culeums.
00:28
And we are asked to find the electric potential, which is equal to the column constant times the second.
00:35
Source charge divided by the distance from the charge.
00:42
Now, we can solve the potential at x by solving for the electric potential along the two components, which is x and y.
00:54
So let's say along x, there are two charges that contributes to the electric potential here.
01:03
One is from this positive charge here, and the other is from.
01:08
This negative charge here.
01:13
Now to solve for the distance between the negative charge and the point, b is the pythagorean theorem which states that the hypotenus is equal to the squared of the two sides, which is four squared plus three squared.
01:35
This gives us a distance between the negative charge and the point of 5 .0.
01:43
Now, the potential along the x is a combination of the potential from the positive charge minus the negative charge.
01:56
Since this is a negative charge, its tendency is to attract a positive test charge.
02:07
And expanding these terms here, we have k times 2 .0 nanoculum divided by 4 .0...