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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 |
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Answer» `E_1:E_2:E_3 = 2/3 :1:2/3` Plate `2` acquires a net positive CHARGE and `3` acquiresnet 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`. 
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