00:01
Problem 23 .32.
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We have a cylinder of the radius of 4 centimeters who's got a non -uniform charge density that depends on the radial distance from its axis.
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It's quadratic in r.
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So this is what it is between r equals 0 and 4 centimeters, and obviously it's zero outside of the cylinder.
00:34
So we want to use gauss's law for this.
00:37
So obviously the charge that's enclosed is going to be row dv, and dv is going to depend on the type of gaussian surface we choose.
00:58
In this case, it's going to be a cylinder.
01:06
So the volume of our cylinder is its circumference times its length, which will be a capital l, and we will hope that that will be irrelevant because it will cancel out later.
01:23
And if not then we're going to have to rethink rethink what we're doing so here's the circumference here's the length there's the radius and so this is our dv so we simply evaluate this and we get pi over 2 a l r to the 4th up to r equals 4 centimeters beyond which this no longer increases so, now to find the electric fields, you know the flux divided by the area, well no, let's do it this way.
02:26
The flux is going to be the electric field times the surface area of a bit of the cylinder we're looking at...