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

[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.

 

[andrewkirk]

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 $a$.

 

[vcsharp2003]

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?

 

[andrewkirk]

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.

 

[vcsharp2003]

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.