Irodov 3.29: A Uniform Field Inside the Hole – Amit & Deepali Sinha Foundation

Problem Irodov 3.29

An infinitely long round cylinder carries charge spread uniformly through its volume with density $\rho$ . Inside it runs a long round cylindrical hole, free of charge, whose axis is parallel to the cylinder’s axis at a distance $a$ from it. Take the permittivity everywhere as that of vacuum. Find the electric field vector $\vect{E}$ inside the hole.

The cylinder in cross-section: charge density ρ fills it except for the cavity, whose axis O' lies a distance a from the cylinder's axis O. — The cylinder in cross-section: charge density $\rho$ fills it except for the cavity, whose axis $O'$ lies a distance $a$ from the cylinder’s axis $O$ .

Given

$\rho$ , uniform
axes a distance $a$ apart
$\varepsilon = 1$

Find

the field

\vect{E}

inside the cavity

Solution

Step 1The field inside a solid charged cylinder

Take first a complete cylinder with no cavity. By symmetry its field is radial and the same all round a coaxial cylinder of radius $r$ and length $\ell$ , so Gauss’s law reads

E\cdot 2\pi r\ell = \frac{\rho\,\pi r^2\ell}{\varepsilon_0}

With $\vect{r}$ the vector drawn to the field point perpendicular from the axis, the field inside is therefore

\vect{E} = \frac{\rho}{2\varepsilon_0}\,\vect{r} \tag{1}

Step 2The cavity as a cylinder of opposite charge

The charged body with its cavity is the same as a full cylinder of density $\rho$ plus a cylinder of density $-\rho$ filling the cavity: inside the cavity the two cancel to zero charge, as they must. Let a point of the cavity sit at $\vect{r}$ from the axis $O$ and at $\vect{r}\,'$ from the cavity’s axis $O'$ , and let $\vect{a}$ run from $O$ to $O'$ .

\vect{r} = \vect{a} + \vect{r}\,' \tag{2}

Applying (1) to each cylinder about its own axis and adding the two fields:

\begin{aligned} \vect{E} &= \frac{\rho}{2\varepsilon_0}\,\vect{r} + \frac{(-\rho)}{2\varepsilon_0}\,\vect{r}\,'\\ &= \frac{\rho}{2\varepsilon_0}\,(\vect{r} - \vect{r}\,') \end{aligned}

and by (2) the bracket is $\vect{a}$ :

\vect{E} = \frac{\rho\,\vect{a}}{2\varepsilon_0}

(3)

The field in the cavity is uniform. For $\rho > 0$ it points along $\vect{a}$ , from the cylinder’s axis toward the cavity’s axis, whatever the point and whatever the two radii.

The cavity, read as a -ρ cylinder superposed on a full +ρ one, and a from O to O'. The field computed at two points of the cavity has the same length and the direction of a (drawn for ρ > 0). — The cavity, read as a $-\rho$ cylinder superposed on a full $+\rho$ one, and $\vect{a}$ from $O$ to $O'$ . The field computed at two points of the cavity has the same length and the direction of $\vect{a}$ (drawn for $\rho > 0$ ).

Check

Two special cases. With $a = 0$ the cavity is coaxial, and the field inside a uniformly charged coaxial tube must vanish by Gauss’s law (a coaxial cylinder inside the cavity encloses no charge); the answer gives $\vect{E} = 0$ . At the cavity’s own axis, $\vect{r}\,' = 0$ , the cavity’s $-\rho$ cylinder contributes nothing, so the field there is the solid cylinder’s (1) at $\vect{r} = \vect{a}$ , which is $\rho\vect{a}/2\varepsilon_0$ , the same value. Units: $\rho a/\varepsilon_0$ is charge per area over $\varepsilon_0$ , a field.

Remark

The same superposition in a ball gives a uniform field $\rho\vect{a}/3\varepsilon_0$ in a spherical cavity: only the coefficient changes, from the cylinder’s $1/2$ to the sphere’s $1/3$ , the reciprocal of the number of dimensions across which the field spreads. Any body whose inside field is linear in position does the same, so an off-centre hole gives a uniform field in an enclosed region without any plates.