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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......... Thanks a lot 😎😎😎😎😎

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.........
           Thanks a lot 😎😎😎😎😎

Grade:11

1 Answers

Arun
25763 Points
2 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

σ=Q/2A

When the plates are brought close at a distance between the two d, its capacitance

C=εr ε0 A/d

where εr is the dielectric constant of the material between the plates (for vacuum, εr =1) and ε0 is the electric constant.

In our case

C=ε0 A/d

We know that electric field due to surface charge distribution σ on a conductor is given by

E =σ/ε0 r
where r 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 = (σ1/ε0 -σ2/ε0) r ….(1)

Here r is unit vector defined by r1- r2

Electric field between parallel plates is also given as

E=V/d r …..(2)
where V is potential difference between the two plates

Comparing (1) and (2) we get

V/d=(Q1-Q2)/(2Aε0)

or V=(Q1-Q2)d/(2Aε0)

Using value of C we get

V=(Q1−Q2)/2C

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