Irodov 3.66: The Floating Pair Splits the Voltage – Amit & Deepali Sinha Foundation

Problem Irodov 3.66

Four large parallel metal plates are stacked with the same small spacing $d$ between neighbors; number them 1 to 4 from the top. The two outer plates, 1 and 4, are joined by a wire, and a potential difference $\Delta\varphi$ is applied between the two inner plates, plate 2 being positive and plate 3 negative. Find:

the electric field strength in each of the three gaps between neighboring plates;
the total charge per unit area on each plate.

Four plates numbered from the top. The outer plates 1 and 4 are joined round the left; the inner plates 2 and 3 are led to terminals, + and -, across which Δφ is applied. — Four plates numbered from the top. The outer plates 1 and 4 are joined round the left; the inner plates 2 and 3 are led to terminals, $+$ and $-$ , across which $\Delta\varphi$ is applied.

Given

4 large plates
spacing $d$
plates 1 and 4 joined
$\Delta\varphi$ between plate 2 ( $+$ ) and plate 3 ( $-$ )

Find

(a)

E

in each gap; (b) the charge per unit area

\sigma_1, \dots, \sigma_4

of each plate

Solution

Step 1Uniform fields and one unknown potential

The plates are large and close, so the field in each gap is uniform and normal to the plates, and equals the potential drop across the gap divided by $d$ . Take plate 3 at zero and plate 2 at $\Delta\varphi$ ; the joined plates 1 and 4 share an unknown potential $\varphi_0$ .

E_{12} = \frac{\Delta\varphi - \varphi_0}{d}, \qquad E_{23} = \frac{\Delta\varphi}{d}, \qquad E_{34} = \frac{\varphi_0}{d}

The signs count $E_{12}$ positive from plate 2 toward plate 1, $E_{23}$ from plate 2 toward plate 3, and $E_{34}$ from plate 4 toward plate 3.

Step 2The joined pair is neutral

The source of $\Delta\varphi$ only moves charge from plate 3 to plate 2, so plates 1 and 4 together keep zero charge, and the system as a whole is neutral, which leaves no field outside the stack and no charge on the two outermost faces. By Gauss’s law a face carries $\varepsilon_0$ times the field leaving it, so plate 1 holds $-\varepsilon_0 E_{12}$ on its lower face and plate 4 holds $+\varepsilon_0 E_{34}$ on its upper face. Their sum is zero:

\begin{aligned} -\varepsilon_0\,\frac{\Delta\varphi - \varphi_0}{d} + \varepsilon_0\,\frac{\varphi_0}{d} &= 0\\ \varphi_0 &= \frac{\Delta\varphi}{2} \end{aligned}

Step 3(a) The fields in the gaps

Substituting $\varphi_0 = \Delta\varphi/2$ into the three gap fields:

E_{23} = \dfrac{\Delta\varphi}{d}, \qquad E_{12} = E_{34} = \dfrac{\Delta\varphi}{2d}

(1)

The inner gap carries the full potential difference; each outer gap carries half of it, and the field points from plate 2 toward plate 1 and from plate 4 toward plate 3.

The solved stack: the joined plates float at Δφ/2. The inner gap carries twice the field of either outer gap, and each face carries one glyph per line ending on it. — The solved stack: the joined plates float at $\Delta\varphi/2$ . The inner gap carries twice the field of either outer gap, and each face carries one glyph per line ending on it.

Step 4(b) The charge on each plate

Each face carries $\varepsilon_0$ times the field leaving it; a plate’s total is the sum over its two faces. Plate 1 has charge only on its lower face, plate 4 only on its upper face:

\sigma_1 = -\varepsilon_0 E_{12}, \qquad \sigma_4 = +\varepsilon_0 E_{34}

Plate 2 sends field out of both faces, plate 3 receives field on both:

\sigma_2 = \varepsilon_0\left(E_{12} + E_{23}\right), \qquad \sigma_3 = -\varepsilon_0\left(E_{23} + E_{34}\right)

Substituting the fields (1):

\begin{gathered} \sigma_1 = -\sigma_4 = -\dfrac{\varepsilon_0\,\Delta\varphi}{2d}\\[4pt] \sigma_2 = -\sigma_3 = \dfrac{3\varepsilon_0\,\Delta\varphi}{2d} \end{gathered}

(2)

Check

The charges in (2) add up as they must. The joined pair is neutral, $\sigma_1 + \sigma_4 = 0$ , and the terminals carry equal and opposite charge, $\sigma_2 + \sigma_3 = 0$ . Going round from plate 2 to plate 3 through plate 1, the wire and plate 4 gives the same potential drop as crossing the inner gap directly:

E_{12}\,d + E_{34}\,d = \frac{\Delta\varphi}{2} + \frac{\Delta\varphi}{2} = \Delta\varphi = E_{23}\,d

Remark

Seen from the terminals the stack is a capacitor of $\sigma_2/\Delta\varphi = 3\varepsilon_0/(2d)$ per unit area: the inner gap, $\varepsilon_0/d$ , in parallel with the two outer gaps in series, $\varepsilon_0/(2d)$ . Nothing of this reaches the outside: with the stack neutral as a whole, the outermost faces carry no charge and the field is confined to the three gaps.