Problem 3.1.3: Circular disk source

Nothing about the disk needs a new idea. Cut it into thin rings (Figure 3.1.27). Each ring is a problem we have already solved, and superposition adds them up. The ring between ${\rho }^{\prime }$ and ${\rho }^{\prime }+d{\rho }^{\prime }$ has length $2\pi {\rho }^{\prime }$ and width $d{\rho }^{\prime }$, so it carries the charge $dQ={\rho }_S(2\pi {\rho }^{\prime })d{\rho }^{\prime }$. It acts like a ring of radius ${\rho }^{\prime }$ with line density ${\rho }_L={\rho }_{S}d{\rho }^{\prime }$. By the result of Problem 3.1.2 with $b\to {\rho }^{\prime }$ and ${\rho }_L\to {\rho }_{S}d{\rho }^{\prime }$, its field on the axis is

and the whole disk is one integral in ${\rho }^{\prime }$ from $0$ to $b$. The same trick builds a solid cylinder from disks and a sphere from rings: once a simple source is solved, it becomes a building block. This is the superposition principle used twice in a row. Let us nevertheless also carry out the full calculation directly from the definition, as a check.

As in Problem 3.1.2, with Q$(x^{\prime },y^{\prime },0)$ now anywhere on the disk,

In polar coordinates $x^{\prime }={\rho }^{\prime }\cos {\varphi }^{\prime }$, $y^{\prime }={\rho }^{\prime }\sin {\varphi }^{\prime }$, so that

With $d{S}^{\prime }={\rho }^{\prime }d{\varphi }^{\prime }d{\rho }^{\prime }$ (Figure 3.1.15c, and do not forget the factor ${\rho }^{\prime }$) the electric field at P is

that is

Since ${\int }_0^{2\pi }\cos {\varphi }^{\prime }d{\varphi }^{\prime }={\int }_0^{2\pi }\sin {\varphi }^{\prime }d{\varphi }^{\prime }=0$ and ${\int }_0^{2\pi }d{\varphi }^{\prime }=2\pi$,

which is precisely the sum of the ring fields $d{E}_z$ found above. With the first standard integral (3.1.30),

and the result is

• $h\ll b$ (very close to the disk, or a very large disk): the bracket tends to $1$ and

  $$E\to \frac{{\rho }_S}{2{\epsilon }_0}{a}_z,$$

  independent of $h$ and of $b$. This is the field of an infinite charged sheet, and letting $b\to \infty$ here is the cleanest derivation of the sheet field we have. It does not decrease with distance at all: the farther we go, the larger the part of the sheet that contributes at a given angle, and the two effects cancel exactly.

• $h\gg b$ (far away): with $h/\sqrt{b^2+h^2}=(1+b^2/h^2{)}^{-1/2}\simeq 1-b^2/(2h^2)$ the bracket tends to $b^2/(2h^2)$, and

  $$\begin{array}{rl}E& \to \frac{{\rho }_S{b}^2}{4{\epsilon }_0{h}^2}{a}_z=\frac{Q}{4\pi {\epsilon }_0{h}^2}{a}_z,\\ Q& =\pi b^2{\rho }_S:\end{array}$$

  a point charge carrying the total charge, once again.

The curve of Figure 3.1.28 shows exactly where each approximation stops being good. For points below the disk ($h<0$) the symmetry of the disk reverses the field. The general formula is $E_z=({\rho }_S/2{\epsilon }_0)[h/|h|-h/\sqrt{b^2+h^2}]$, and crossing the disk at its centre the field jumps from $-{\rho }_S/(2{\epsilon }_0)$ to $+{\rho }_S/(2{\epsilon }_0)$, a jump of ${\rho }_S/{\epsilon }_0$ caused by the surface charge.

Units: ${\rho }_S/{\epsilon }_0$ is in V/m and the bracket is a pure number. Both routes give the same integral. The two limits reproduce the infinite sheet and the point charge, the two results we expect.