Four large metal plates are located a small distance apart from one another as shown in the Figure. The extreme plates are connected by means of a conductor, while a potential difference V is applied to the internal plates. Find the electric fields between the neighbouring plates.

$E_{23}=\frac{V}{d}, E_{34}=\frac{V}{2 d}$

We may take the plates to be connected as shown in the Figure.

Let E₂₃ be the electric field between plates 2 and 3. Then from the fact that electric field = rate of change of potential, $E_{22}=\frac{V}{d}$.

Applying `loop rule' to the loop

`A – 1 – 2 – 3 – 4 – B – A'

V' – V – + V' ⇒ V' = $\frac{V}{2}$

where V' is the numerical value of the potential difference between plates 1 and 2 or 3 and 4

∴ $E_{12}$  or  $E_{34} = \frac{V}{2 d}$

$E = \frac{\sigma}{\varepsilon_0}$

∴ $\sigma_1$ (density of charge on the plate 1)

$=\varepsilon_0 E_{12}=\varepsilon_0 \frac{V}{2 d}=\sigma_4$  (numerically)

The surface density of charge of the right surface of plate 2 is $\sigma'=\varepsilon_0, E_{23}=\varepsilon_0 V d$

∴ $\sigma_2$ (surface density of charge on plate 2)

$=\frac{\varepsilon_0 V}{2 d}+\frac{\varepsilon_0 V}{d}=\frac{3}{2} \frac{\varepsilon_0 V}{d}$

Thus

(i) $E_{23}=\frac{V}{d} E_{12}$ or $E_{34}=\frac{V}{2 d}$

(ii) $\sigma_1=\sigma_4=\varepsilon_0 \frac{V}{2 d} \sigma_2=\varepsilon_0=\frac{3}{2} \varepsilon_0 \frac{V}{d}$.