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Charge Q is given to metal plate of surface area A. It gets equally divided on the total surface, both sides. As such surface charge density on each side is
ฯ=Q/2A
When the plates are brought close at a distance between the two d, its capacitance
C=ฮตr ฮต0 A/d
where ฮตr is the dielectric constant of the material between the plates (for vacuum, ฮตr =1) and ฮต0 is the electric constant.
In our case
C=ฮต0 A/d
We know that electric field due to surface charge distribution ฯ on a conductor is given by
Eย =ฯ/ฮต0ย rwhereย rย is unit vector perpendicular to the conductor's surface.
Since both charges are of same kind net electric field between the plates is difference of two fields.
E1-E2 = (ฯ1/ฮต0 -ฯ2/ฮต0)ย rย โฆ.(1)
Hereย rย is unit vector defined byย r1-ย r2
Electric field between parallel plates is also given as
E=V/dย rย โฆ..(2)whereย Vย is potential difference between the two plates
Comparing (1) and (2) we get
V/d=(Q1-Q2)/(2Aฮต0)
orย V=(Q1-Q2)d/(2Aฮต0)
Using value of C we get
V=(Q1โQ2)/2C
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