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 plate is

Options

A. `E_1:E_2:E_3 = 2/3 :1:2/3`
B. `E_1:E_2:E_3 = 1/2 : 1: 1/2`
C. and variation of electric potential will be

D.

Answer

B, D

Explanation

b.,d.
Plate `2` acquires a net positive charge and `3` acquires net negatie charge.
Due to induction, plate `1` acqurires negative charge and plate `4` positive charge. `1` and `4` are equipotential (since they are joined).
Due to symmetry
`V_(1-2)=V_(3-4)`
and `V_(2-3)=2V_(2-1)=2V_(4-3)`
because charge capacitor `2-3` is double.
Electric field `=("potential")/(d)`
`E_(1):E_(2):E_(3)=1:2:1=(1)/(2):1:(1)/(2)`
and form 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 directon of `E`.