Electric field at the centre of a charged disc of uniform charge density

The electric field at any point at a finite distance z from the centre of a charged disc of uniform charge density along the axis of the disc is given by the equation: $$E= \int dE=\dfrac{\sigma z}{4\epsilon_0} \int_0^R (z^2 +r^2)^{-3/2}(2r) dr = \dfrac{\sigma}{2\epsilon_0}\left(1 - \dfrac{z}{\sqrt{z^2 + R^2}}\right)$$

According to this equation, at $z=0$, the electric field is :$$E = \dfrac{\sigma}{2\epsilon_0}$$ However, logically, I feel that at the centre of the disc, the field must be zero as all the field vectors get cancelled at that point. Why does this happen?

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Your equation is valid for z>0 – yankeefan11 Jul 17 '13 at 16:00
yeah i know. for a general solution, we need to change the equation by replacing z with |z| and multiplying the entire equation with z/|z| to get the correct direction of the field vector, but it still doesnt answer my question as to what will be the electric field at the centre of the disk. – Ajaykrishnan Jayagopal Jul 17 '13 at 16:13
check this link: <web.mit.edu/viz/EM/visualizations/coursenotes/modules/…; Scroll down to the section regarding uniform charge distributions. it gives the solution for z>0 and for z<0 but it does not tell what happens at z=0. The graph shown also describes z=0 as a point of discontinuity. – Ajaykrishnan Jayagopal Jul 17 '13 at 16:16
Yeah. The E field will be canceled at the center. It will also change direction, thus a discontinuity – yankeefan11 Jul 17 '13 at 16:31

Like James Maslek said, this is just an effect of having an infinitely thin disk--the field is a step function.

If you like, you can replace your disk with two disks , each having the same radius and surface charge density $\sigma/2$. Seperate them by some finite distance, so they are located at $z = \pm \epsilon$ for some small $\epsilon$. Then, work out what the field is at some arbitrary point along their common axis. You will find that the field at $z=0$ is zero for all $\epsilon$ and that your solution limits to your original equation for $z \gg \epsilon$. This way, you can ignore the discontinuous nastiness of the infinitely thin disk.

If you want to get yet fancier, and REALLY analyze this situation in the case where the thickness of the disk is important, you can think of a cylinder with uniform volume charge density, radius $r$ and thickness $L$. You can work out the field at an aribtrary point along the cylinder's axis, inside or outside. It should be easy enough to show that the field at the center of the cylinder is zero. Then, you can replace $\rho$ with $\frac{\sigma}{L}$ and see how it compares with your expression for a disk in the limit $z \gg L$, which should once again equal your original expression.

Finally, a third way to see this, but one which involves much harder calculus would be to derive the field due to a disk at an arbitrary point on the plane of the disk, and then see that the field goes to zero at the center of the disk when you are constrained to the plane.

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Consider the charged disk. Every dE you have will cancel out the dE opposite to it. This is because of the symmetry to the problem.

Another way to view this is that the field points one direction above the disk and then a different direction on the other side. This is essentially a step function. You have a discontinuity at the center (z=0), therefore your result for E(x=0) cannot be described using your function above. You have to use the symmetry of the problem. Because of the uniform distribution, everything cancels at the center

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