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Consider a thin hollow cylindrical insulating shell like the one above draw in 3D-coordinate with Z-axis going out of page and is centered at the origin of XYZ axis. It has electric charge +Q uniformly distributed with radius R and length L and the shell's ends are open.

  1. The question asks me to find the direction of electric field at the origin(at the center of the cylinder).

However, I can't even figure out where to begin. If there were charges floating around in the cylinder, wouldn't they just point radially outward? How could they all align to point in one direction? the options are each axis in positive negative direction or 0.

  1. the question asks if the electric field will increase as it goes from -L/2 to L/2 on x-axis.

I'd assume the electric field would increase til the origin and hit 0 at the origin due to symmetry and increase back again as it moves further away from the origin.

  1. the question asks for V with respect to distance like in #2.

I think this will be just the graph from #2 flipped around x-axis since V=-E*distance.

Any help will be greatly appreciated.


1 Answer 1


The description of the system is not very clear. I'm assuming that the x-axis is the symmetry axis of the cylinder and that the cylinder extends from x=-L/2 to x=L/2. I'm also assuming that there is a uniform surface charge density on the shell, not a uniform volume charge density over the entire cylinder.

Q1 hint: use symmetry arguments for the direction and magnitude of the field.

Q2 hint: in your wording, I think you meant 'decrease' for the first occurence of 'increase'. Otherwise, your approach looks reasonable; maybe the question expected you to consider negative fields rather than just the magnitudes of the field.

Q3 hint: This is hard work; the easiest approach is to write down an expression for the potential at point $x$ from a ring-shaped charge at point $x'$, and then integrate.

  • $\begingroup$ Yes, that's exactly how the system is set up. For Q1 I'm assuming the answer is 0 since the electric field from floating charges eventually cancels out at the origin. $\endgroup$
    – Hello
    May 28, 2016 at 10:57

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