# Two identical metal plates one having Q1 charge and another Q2 [Q1>Q2] are given. A parallel plate air capacitor of capacitance 'C' is formed by bringing them together. Determine the potential difference between the plates.Please solve the question immediately.........ย  ย  ย  ย  ย  ย Thanks a lot ๐๐๐๐๐

Arun
25750 Points
5 years ago

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 r
whereย 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 = (ฯ1/ฮต0 -ฯ2/ฮต0)ย rย โฆ.(1)

Hereย rย is unit vector defined byย r1-ย r2

Electric field between parallel plates is also given as

E=V/dย rย โฆ..(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=(Q1โQ2)/2C