It seems that most people agree to the field inside a hollow cylinder being zero but I’m troubled when I consider the case of a ring shaped linear distribution of charge.

It is known that the electric field due to a ring charge distribution inside it(in the plane) is non zero (don’t take my word for it, https://youtu.be/BF3wEV4tWq8)

Suppose we consider the ring shaped lamina of the cylinder that lies in the horizontal plane perpendicular to the axis of the infinite cylinder. The field due to it would be non-zero and radially inward as Ive said above, but now we consider the rest of the cylinder. Now, In order for the field inside to be zero, it would mean that the field caused due to the rest is in the opposite direction to that of the ring I.e radially outward! This seems very counterintuitive to me. Will it not be in the same direction as caused by the ring?

Is there anywhere I’m going wrong?

  • $\begingroup$ It would help if you write the ring solution down in the question $\endgroup$
    – JEB
    Commented Jul 17, 2020 at 2:31
  • $\begingroup$ @JEB well actually there wasn’t an exact solution to the ring’s field, it’s a tricky integral, but we can evaluate whether it’ll be positive or negative I.e inward or outward $\endgroup$ Commented Jul 17, 2020 at 2:33
  • $\begingroup$ @JEB but even if we didn’t know the field due to a ring, wouldn’t the rest of the cylinder have a field that’s in the same direction as well? $\endgroup$ Commented Jul 17, 2020 at 2:34

1 Answer 1


The field inside a cylinder is zero, and you can make symmetry arguments that make that clear.

The field inside a flat ring ($R$) is given by:

$$ E\big(\frac {\rho} R\big)= \int_0^{2\pi}\frac{\cos\theta-\frac {\rho} R}{[1+(\frac {\rho} R)^2-\frac{2{\rho}}R\cos\theta]^{\frac 3 2}}$$

which is not zero. How does that work out?

If you look at the divergence in cylindrical coordinates, inside the cylinder/ring:

$$ \vec{\nabla}\cdot \vec E =\frac 1 {\rho}\frac{\partial (\rho E_{\rho})}{\partial \rho } + \frac 1 {\rho}\frac{\partial E_{\phi}}{\partial \phi} + \frac{\partial E_z}{\partial z} =0$$

In the cylindrical case, $E_{\phi}=E_z=0$, by symmetry, so the radial field must go like:

$$ E_{\rho}(\rho) = \alpha \times \frac 1 {\rho}, $$

but that has a non-zero divergence at the origin, hence the constant $\alpha=0$.

In the ring case, field can "leak out" in the $z$-direction so that even though $E_z(z=0)=0$,

$$ \frac{\partial E_z}{\partial z}_{|_{z=0}} \ne 0 $$

The divergence from the $z$ component exactly cancels the radial divergence.

Now consider the direction of the field, as discussed at 1:00 in the video. We know it must point inward (for positive charge) because as $r \rightarrow R$, the ring appears as a line charge and we must get the usual $1/r$ behavior in close proximity.

The point is, the field their is completely dominated by the nearby infinitesimal line element of the ring.

Now consider the cylinder minus the ring (or a small band, if need be): The nearest pieces of charge are not in the $\rho$ direction, rather, there are equal pieces in the $\pm z$ directions, and their fields nearly cancel. Meanwhile, the charges at the opposite side of configuration are pushing (a fiducial $+$ charge) both in the same direction. Thus that field points outward.

So as you suspected, the field inside the cylinder of a cylinder minus a central band is exactly opposite the field from the band alone.


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