Irodov 3.67: Smear the Charge into a Sheet – Amit & Deepali Sinha Foundation

Problem Irodov 3.67

Two infinite parallel conducting plates, labelled 1 and 2, are a distance $l$ apart. A point charge $q$ is placed between them, a distance $x$ from plate 1. Find the charge induced on each plate.

The two infinite conducting plates, a distance l apart, and the point charge q between them, a distance x from plate 1. — The two infinite conducting plates, a distance $l$ apart, and the point charge $q$ between them, a distance $x$ from plate 1.

Given

plate separation $l$
point charge $q$
its distance $x$ from plate 1
plates 1 and 2 earthed (assumed)

Find

the induced charges

q_1

and

q_2

Solution

Step 1The induced totals do not care where the charge sits sideways

Both plates are assumed earthed, so they share one potential, which is what makes the induced charge a definite response to $q$ alone. Slide $q$ parallel to the plates and the whole picture slides with it, so the charge it induces on each plate keeps the same total. By superposition, replacing $q$ with many small charges spread over the plane through $q$ , adding up to $q$ , therefore leaves both totals $q_1$ and $q_2$ unchanged; only their spread changes.

Step 2The fields of the smeared sheet

q smeared into a uniform sheet σ at its height: the field E_1 above it and E_2 below it, uniform and drawn to scale for x = 0.35 l, and the induced charge on the facing surfaces, denser where the field is stronger. — $q$ smeared into a uniform sheet $\sigma$ at its height: the field $\vect{E}_1$ above it and $\vect{E}_2$ below it, uniform and drawn to scale for $x = 0.35\,l$ , and the induced charge on the facing surfaces, denser where the field is stronger.

With the charge spread at density $\sigma = q/S$ over a large area $S$ , the field is uniform on each side of the sheet and normal to it: $E_1$ toward plate 1, $E_2$ toward plate 2. Gauss’s law across the sheet fixes their sum:

E_1 + E_2 = \frac{\sigma}{\varepsilon_0} \tag{1}

The plates are at one potential, so the potential falls by the same amount from the sheet to either plate:

E_1\,x = E_2\,(l - x) \tag{2}

Eliminating $E_2$ from (1) with (2):

\begin{aligned} E_1 + E_1\,\frac{x}{l - x} &= \frac{\sigma}{\varepsilon_0}\\ E_1\,\frac{l}{l - x} &= \frac{\sigma}{\varepsilon_0} \end{aligned}

so that, with $E_2$ back from (2),

E_1 =\frac{\sigma}{\varepsilon_0}\,\frac{l - x}{l}, \qquad E_2 = \frac{\sigma}{\varepsilon_0}\,\frac{x}{l}

Step 3The induced charges

The field ends on the facing surface of each plate, whose charge density is $-\varepsilon_0$ times the field arriving there; beyond the plates there is no field. Multiplying by the area $S$ , with $\sigma S = q$ :

q_1 = -\varepsilon_0 E_1 S = -\sigma S\,\frac{l - x}{l}, \qquad q_2 = -\varepsilon_0 E_2 S = -\sigma S\,\frac{x}{l}

q_1 = -q\,\dfrac{l - x}{l}, \qquad q_2 = -q\,\dfrac{x}{l}

(3)

Check

Every line of $q$ ends on one plate or the other, and (3) gives exactly that:

q_1 + q_2 = -q\,\frac{(l - x) + x}{l} = -q

Bring the charge up to plate 1, $x \to 0$ : then $q_1 \to -q$ and $q_2 \to 0$ , the whole induced charge gathering on the plate it touches. At the midpoint, $x = l/2$ , the two plates share $-q/2$ each, as symmetry requires.

Remark

The share of plate 1 is the linear interpolation $(l - x)/l$ : it is the potential that the point where $q$ sits would have if plate 1 were held at unit potential and plate 2 at zero, with $q$ removed. That is Green’s reciprocity theorem at work, and it works for any shape of earthed conductors: the charge a point charge induces on one of them is $-q$ times that potential, which is why smearing $q$ along an equipotential plane could not change the answer.