# Why is electric field at the center of a charged disk not zero?

• vcsharp2003
In summary, according to symmetry of concentric rings about the center, the electric field at the center of a uniformly charged disk should be zero, but this is not the case due to the finite contribution of each concentric ring.
vcsharp2003
Homework Statement
A disk that is very thin and uniformly charged has a non-zero electric field at its center. Explain why this statement is true or false.
Relevant Equations
##E = \dfrac {\sigma} {2 \epsilon_{0}} [1 - \dfrac {x} {(R^2 + x^2)^{\frac {1}{2}}} ]##, where
##E## is electric field strength at a point P on the axis of the disk
##x## is distance of point P from the center of the sphere ( axis is perpendicular to the disk and passing through its center)
##R## is radius of disk
##\sigma## is the uniform charge density per unit area of the thin disk
##\epsilon_{0}## is electrical permittivity of vacuum
The electric field strength at the center of a uniformly charged disk should be zero according to symmetry of concentric rings about the center, where each ring is contributing to the electric field at the center of the disk.

For a thin ring of uniform charge distribution the formula is ##E = \dfrac {1} {4 \pi \epsilon_{0}} \dfrac {Qx} {(R^2 + x^2)^{\frac {3}{2}}}##, where the electric field ##E## is at a point P that is a distance ##x## from the center of the ring and along the ring's axis. When we consider the center of the ring, then ##x =0## which gives us ##E = 0## at the center of the ring.
Thus, each concentric ring will contribute ##0## to the electric field at the center of the thin disk. Consequently, the electric field at the center of the thin disk must be ##0##.

However, I do see from the formula for a thin charged disk as given under the relevant equations, that the electric field at the center of a disk is found to be ##E = \dfrac {\sigma} {2 \epsilon_{0}}## when we substitute ##x =0## in the mentioned formula.

I am unable to understand the flaw in my logic.

Last edited:
Your logic is correct. The field will be zero.
The formula you quote only holds for ##x\neq 0##. It does not apply when ##x=0##. For an infinitely thin disc, the electric field is discontinuous at the centre of the disc. In practice, no disc is infinitely thin so we don't need to worry about that.

The reason the formula does not apply when ##x=0## is that we obtain it by integrating the formula for field from an infinitesimal ring, for ring radius ##a\in [0,R]##. The integrand for that is always finite if ##x>0## but if ##x=0## it becomes infinite at ##a=0## and the integral diverges. So the formula's derivation is invalid if ##x=0##. See here for a derivation of the formula for the disc. You can see how they use that integral over ##a##.

MatinSAR, PhDeezNutz and vcsharp2003
andrewkirk said:
The formula you quote only holds for x≠0. It does not apply when x=0. For an infinitely thin disc, the electric field is discontinuous at the centre of the disc. In practice, no disc is infinitely thin so we don't need to worry about that.
Ok. Thanks for the excellent explanation.
It seems that if we plot a graph of ##E## vs ##x## for a thin charged disk then the the y-axis will be a vertical asymptote to the plot. Is that correct?

vcsharp2003 said:
Ok. Thanks for the excellent explanation.
It seems that if we plot a graph of ##E## vs ##x## for a thin charged disk then the the y-axis will be a vertical asymptote to the plot. Is that correct?
It will have a stranger form than that: no asymptote. The function will tend to a finite nonzero limit, per the above formula, as ##x\to 0##. But the actual value of the function at ##x=0## will be 0, which is different from the limit. Effectively, it is the function specified by
$$f(x) = \left\{ \begin{array}{ll} -\dfrac {\sigma} {2 \epsilon_{0}} [1 - \dfrac {x} {(R^2 + x^2)^{\frac {1}{2}}} ] & x < 0 \\ 0 & x = 0\\ \dfrac {\sigma} {2 \epsilon_{0}} [1 - \dfrac {x} {(R^2 + x^2)^{\frac {1}{2}}} ] & x > 0\\ \end{array} \right.$$
Note that not only is the function at 0 not equal to the limit, but the left and right limits also differ.

Last edited:
vcsharp2003
andrewkirk said:
It will have a stranger form than that: no asymptote. The function will tend to a finite nonzero limit, per the above formula, but with ##x=0##, as ##x\to 0##. But the actual value of the function at ##x=0## will be 0, which is different from the limit. Effectively, it is the function specified by
$$f(x) = \left\{ \begin{array}{ll} -\dfrac {\sigma} {2 \epsilon_{0}} [1 - \dfrac {x} {(R^2 + x^2)^{\frac {1}{2}}} ] & x < 0 \\ 0 & x = 0\\ \dfrac {\sigma} {2 \epsilon_{0}} [1 - \dfrac {x} {(R^2 + x^2)^{\frac {1}{2}}} ] & x > 0\\ \end{array} \right.$$
Note that not only is the function at 0 not equal to the limit, but the left and right limits also differ.
That is very beautifully explained. It makes complete sense now. Thankyou.

PhDeezNutz

## 1. Why is the electric field at the center of a charged disk not zero?

The electric field at the center of a charged disk is not zero because of the distribution of charge on the disk. The charge on the disk is not uniformly distributed, meaning that there are areas of higher and lower charge density. This results in an uneven distribution of electric field lines, with some lines pointing towards the center of the disk and others pointing away from it. As a result, the net electric field at the center is not zero.

## 2. How does the size of the charged disk affect the electric field at its center?

The size of the charged disk does not have a significant effect on the electric field at its center. As long as the charge distribution on the disk remains the same, the electric field at the center will also remain the same. However, if the size of the disk is increased or decreased, the overall strength of the electric field may change due to the change in the total charge on the disk.

## 3. Is the electric field at the center of a charged disk always in the same direction?

No, the electric field at the center of a charged disk can vary in direction depending on the distribution of charge on the disk. If the charge is evenly distributed, the electric field lines will be symmetric and point directly towards or away from the center. However, if the charge is unevenly distributed, the electric field lines may point in different directions, resulting in a net electric field that is not in a specific direction.

## 4. Can the electric field at the center of a charged disk be zero under any circumstances?

Yes, the electric field at the center of a charged disk can be zero if the charge is evenly distributed on the disk and the disk is perfectly symmetrical. In this case, the electric field lines will cancel each other out, resulting in a net electric field of zero at the center.

## 5. How does the distance from the center of the disk affect the strength of the electric field?

The distance from the center of the disk does not have a direct effect on the strength of the electric field. However, as you move further away from the center, the electric field lines become more spread out, resulting in a weaker electric field. This is due to the inverse square law, which states that the strength of the electric field decreases with the square of the distance from the source of the field.

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