Four identical metallic plates (1,2,3 and 4) are arranged in air at same distance d from each other with their outer plates being connected together and earthed. If the plates 2 and 3 are connected with a cell of constant emf E, then ratio of electric fields between the plates is

Plate 2 acquires a net positive charge and 3 acquires a net negative charge.

Due to induction, plate 1 acquires negative charge and plate 4 positive charge. 1 and 4 are equipotential (since they are joined). 

Due to symmetry

${\mathrm{V}}_{1-2}={\mathrm{V}}_{3-4}$

and ${\mathrm{V}}_{2-3}=2{\mathrm{V}}_{2-1}=2{\mathrm{V}}_{4-3}$

because charge capacitor 2-3 is double.

$\text{ Electric field }=\frac{\text{ potential }}{\mathrm{d}}$

${\mathrm{E}}_1:{\mathrm{E}}_2:{\mathrm{E}}_3=1:2:1=\frac{1}{2}:1:\frac{1}{2}$

and from plate 1 to 4

Potential first increases, then decreases, and again increases.

Since we are going first in direction opposite of E and then in direction of E.