Solutions to problem

[00:00:01]Hello everyone. This is Oregon and I will present to you the solution to problem 260 which asks us to deal with an electrically charged charged plastic pipe.

[00:00:15]The problem provides a couple of information.

[00:00:20]We know the thickness of the pipe which is the difference between the two radii B and A.

[00:00:28]We also know that the pipe holds a total charge which is expressed in um coulomb per meters. Is this value lambda?

[00:00:42]I also made a notation. I identify the pipe's length as L.

[00:00:48]Uh I will use this um value but as you will see uh this value will not be part of the final results.

[00:00:58]Because uh this problem feels like a natural continuation of 200 of problem 250 nine I will use the same approach where to make things easier when solving, I will uh calculate the charge density this value row.

[00:01:20]Uh here is the math for that.

[00:01:22]This is the value that I will use.

[00:01:27]We also need to remember Gauss's law which we will solve for a cylinder because the shape of the cylinder provides symmetry so this makes uh easy it makes it easy to solve this integral which states that the net electric flux flux that passes through an enclosed uh surface will will this this dot product will equal the value of the in the the charge that's enclosed by the surface divided by the permittivity of vacuum.

[00:02:18]Now, if we solve this integral, we will get this nice value.

[00:02:25]Uh I will make just a very short comment.

[00:02:29]Uh as we know, the cylindrical uh an enclosed uh surface that shapes as a cylinder also has two additional um areas besides the uh external area.

[00:02:49]Now, this this two surfaces, you will see that are not part of the solution.

[00:02:55]And this the the the very quick uh tip that I will provide is here in inside this integral, we are dealing with a dot product between two vectors.

[00:03:09]The electric field vector and the normal vector to the infinitesimal uh area element.

[00:03:21]Now, when these two vectors are perpendicular, their dot product will equal zero.

[00:03:29]So, with these two uh tools, uh we will jump into solving um the values for the electric field in the two uh sorry, in the three regions uh that we are asked to.

[00:03:47]The first region is uh inside the pipe.

[00:03:51]The second region is inside the thickness of the pipe, and the third region is outside the pipe.

[00:03:58]Now, the first region we are asked to calculate the electric field is a rather trivial question because the surface that the cylinder that's [snorts] inscribed in the in the in the interior of the pipe is enclosing a charge that is null.

[00:04:25]So, of course, the electric field in this case will be zero, also.

[00:04:31]The almost trivial question is this one where we are asked to evaluate the electric field outside the pipe.

[00:04:42]I'm saying it's almost trivial because the enclosed charge is the total amount of the charge that the pipe holds, and this is a given by the problem, so we only need to use this value and make an input in the equation for the electric field as we solved it here.

[00:05:09]And if we do the math, this is the final value that we will get.

[00:05:15]Now, the most involved one uh is this case where we are asked to calculate the electric field inside the thickness of the pipe.

[00:05:27]For that, we will use the charge density that we calculated.

[00:05:32]And with it, we will identify the value for the charge that's enclosed by our surface.

[00:05:43]So, with this value, if we do the math and we will insert this into the equation of the that we got from solving the Gauss's law.

[00:05:57]And if we are patient to properly do the math this is the final that value that we will get.

[00:06:08]Now from what I saw there were quite a lot of people that got this right.

[00:06:15]I'm going to try and check and see everyone that managed to properly and correctly answer this this problem and I will mention them in the comments.

[00:06:32]I will only wish everybody a nice end of the weekend and we will see each other when we will have to deal with the next problem 261.