I have a question that my teacher was explaining but the question seemed to contradict the real world. I know it has got to be my understanding that is flawed but I cannot seem to get it. So here goes: enter image description here

In this question it asks you to get the magnitude of the electric field at a radius of 1cm from the charged cable, and this cable is surrounded by this cylindrical shell which also has charge in it. Then my teacher said to draw a Gaussian surface of radius 1cm to enclose the wire and said that because the only charge inside this surface is of the cable then we ignore the charge of the cylindrical shell.

So my question is how could the cylindrical shell has no effect whatsoever on the point when the shell is in it self charged and the electric field should affect the whole area around it?

  • \$\begingroup\$ why didn't you ask your teacher this question? ... it only makes sense, since the teacher can identify the study materials you may need to review \$\endgroup\$
    – jsotola
    Jul 13, 2022 at 19:35
  • \$\begingroup\$ There is no electric field directed inwards from the outer cylindrical shell <-- physics. It has nowhere to go. \$\endgroup\$
    – Andy aka
    Jul 13, 2022 at 20:16
  • \$\begingroup\$ @jsotola he is unavailable this week \$\endgroup\$ Jul 13, 2022 at 20:21
  • \$\begingroup\$ @Andyaka ok why??? \$\endgroup\$ Jul 13, 2022 at 20:21

3 Answers 3


my question is how could the cylindrical shell has no effect whatsoever on the point when the shell is in it self charged and the electric field should affect the whole area around it?

There are a couple of ways we can understand this.

Let's start with an easy case. Imagine a charge at point P directly at the center of the hollow cylindrical shell. Now imagine a charge at point A on the shell. This charge by itself would indeed exert a force on P. But if the shell is uniformly charged, then there should be another charge at point C which is equal in magnitude. C will also exert a force on A, but in exactly the opposite direction. Therefore, the net force on P is zero.

To calculate the total force at P you could do an integral and add up the forces due to each charge at every point in the shell. I am not going to show the integral here, but as it turns out, you could move P anywhere inside the shell (not just the center) and the result of the integral always works out to be zero.

enter image description here

  • \$\begingroup\$ The OP can just go here for the math. So long as it is a square-law force (and it is) the 'shell theorem' used to show a shell with uniform density (consistent with a charged shell where the charge distributes uniformly) exerts no net force on a mass inside the shell similarly applies. \$\endgroup\$
    – jonk
    Jul 14, 2022 at 0:37

The electric field inside any closed hollow conductor is zero. Assuming there is no charge inside the shell.

To see why the cylinder does not contribute to the electric field, imagine we remove the charged line and just have the shell. Now put a gaussian cylinder inside of the shell.

The electric field inside the shell is zero because there is no charge for the gaussian cylinder to enclose. This leaves the line of charge as the only contributing factor to the electric field inside the shell.

  • \$\begingroup\$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. \$\endgroup\$
    – Community Bot
    Jul 14, 2022 at 4:25

All charges are relative unless there is a reference. The same is true for voltages and that reference = 0V is called "ground" regardless if it floating or if that "ground" is connected to "Protective Earth" ground aka PE gnd.

So the point of this quiz was to choose the shell ( or shield) as 0 V with no contribution or effect from the charge or flow of charges= current

This is the theory, but in practice, the resistance is non-zero so high voltage coaxial cables when used, need to be PE earth grounded frequently along a transmission line.

For a complete analysis.


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