Four large metal plates are located at a small distance `d` from one another as shown in Fig. The exremen plates are interconnected
by means of a conductor while a potential difference `Delta varphi` is applied to inernal plates. Find :
(a) the values of the electric field strength between neighbouring plates,
(b) the total charge per unit area of each plate.

Explanation

(a) As the metallic plates 1 and 4 are located and connected by means of a conductor, `varphi_(1) = varphi_(4)`. Plates 2 and 3 have the same amount of positive and negative charges and due to induction, plates 1 and 4 are respectively negatively and positively charged adn in addition to all the four plates are located a small but at equal distance `d` relative to each other, the magnitude of electric field strength between 1-2 and 3-4 are both equal in magniutude and direction (say `vec(E)`). Let `vec(E)` be the field strength between the plates 2 and 3, which is directed from 2 to 3. Hence `vec(E) uarr darr vec(E)` (Fig).
According to the problem
`E' d = Delta varphi = varphi_(2) - varphi_(3)` ...(1)
In addition to
`varphi_(1) - varphi_(4) = 0 (varphi_(1) - varphi_(2)) + (varphi_(2) - varphi_(3)) + (varphi_(3) - varphi_(4))`
`0 = -Ed + Delta varphi - Ed`
or, `Delta varphi = 2Ed` or `E = (Delta varphi)/(2d)`
Hence `E = (E')/(2) = (Delta varphi)/(2d)` ....(2)
(b) Since `E alpha sigma` we can state that according to equaction (2) for part (a) the change on the plate 2 is divided into two parts, such that `1//3 rd` of it lies on teh upper side and `2//3` on its lower face.
Thus chage density of upper face of plate 2 or of plate 1 or plate 4 and lower face of `3 sigma = epsilon_(0) E = (epsilon_(0) Delta varphi)/(2d)` and charge density of lower face of 2 or upper face of 3
`sigma' = epsilon_(0) E' = epsilon_(0) epsilon_(0) (Delta varphi)/(d)`
Hence the net charge density of plate 2 or 3 becomes `sigma + sigma' = (3 epsilon_(0) Delta varphi)/(2d)`, which is obvious from the argument .