Irodov 3.59: Where the Flux of q Lands – Amit & Deepali Sinha Foundation

Problem Irodov 3.59

A point charge $q$ is at a distance $l$ from an infinite conducting plane. Find the surface density of the charge induced on the plane as a function of the distance $r$ from the foot of the perpendicular dropped from the charge to the plane.

The point charge q at distance l from the conducting plane, the foot O of the perpendicular, and a point P of the plane at distance r from O. — The point charge $q$ at distance $l$ from the conducting plane, the foot $O$ of the perpendicular, and a point $P$ of the plane at distance $r$ from $O$ .

Given

point charge $q$
distance $l$ from the plane

Find

the induced surface density

\sigma(r)

Solution

Step 1The field at the surface from the charge and its image

The plane is an equipotential at zero, so above it the field is that of $q$ and its image $q' = -q$ at the mirror point, a distance $l$ below $O$ . A point $P$ of the plane is the same distance from both.

\rho = \sqrt{l^2 + r^2} \tag{1}

Each charge therefore gives a field of the same size at $P$ .

E_1 = E_2 = \frac{1}{4\pi\varepsilon_0}\frac{q}{\rho^2}

At the point of the plane a distance r from the foot of the perpendicular: the field E_1 of q, the field E_2 of its image q' (dashed), and their sum E normal to the plane. The glyphs show the induced charge, densest under q. — At the point of the plane a distance $r$ from the foot of the perpendicular: the field $\vect E_1$ of $q$ , the field $\vect E_2$ of its image $q'$ (dashed), and their sum $\vect E$ normal to the plane. The glyphs show the induced charge, densest under $q$ .

The components along the plane cancel, and the components normal to it add, each a fraction $l/\rho$ of its field. For $q > 0$ the sum points into the metal.

E = 2\cdot\frac{1}{4\pi\varepsilon_0}\frac{q}{\rho^2}\cdot\frac{l}{\rho} = \frac{ql}{2\pi\varepsilon_0 (l^2 + r^2)^{3/2}} \tag{2}

Step 2The surface density from the normal field

At the surface of a conductor the field is normal and $E_n = \sigma/\varepsilon_0$ , where $E_n$ is the component along the outward normal, from the metal into the vacuum. The field (2) points into the metal, so $E_n = -E$ .

\sigma = \varepsilon_0 E_n = -\varepsilon_0\,\frac{ql}{2\pi\varepsilon_0 (l^2 + r^2)^{3/2}}

\sigma(r) = -\dfrac{ql}{2\pi (l^2 + r^2)^{3/2}}

(3)

The induced density σ against r, for q > 0: negative everywhere, largest in magnitude under the charge (r = 0), and dying away within a few l. — The induced density $\sigma$ against $r$ , for $q > 0$ : negative everywhere, largest in magnitude under the charge ( $r = 0$ ), and dying away within a few $l$ .

The induced charge has the sign opposite to $q$ everywhere, and its magnitude falls to about a third of its peak value at $r = l$ .

Check

Every field line of $q$ ends on the plane (none reaches infinity or enters the metal), so the total induced charge must be $-q$ . Integrating (3) over rings of radius $r$ and width $dr$ :

\int_0^\infty \sigma\,2\pi r\,dr = -ql\int_0^\infty \frac{r\,dr}{(l^2 + r^2)^{3/2}} = -ql\left[-\frac{1}{\sqrt{l^2 + r^2}}\right]_0^\infty = -ql\cdot\frac{1}{l} = -q

as required. Units: $q/l^2$ is a charge per area.

Remark

The same integral taken only out to a radius $R$ gives the induced charge inside it,

q_{\text{in}}(R) = -q\left(1 - \frac{l}{\sqrt{l^2 + R^2}}\right)

which reaches half of $-q$ at the radius

\frac{l}{\sqrt{l^2 + R^2}} = \frac{1}{2} \implies R = \sqrt{3}\,l

Half the induced charge sits within a disc only $\sqrt{3}$ times wider than the height of the charge, but the tail is long: a tenth of it still lies beyond $R \approx 10\,l$ .