# Capacitance of Parallel Plate Capacitor

Choose two parallel conducting plates with charge densities separated by a distance ‘d’ meters as shown in the figure. Assume charges are distributed uniformly on the plates.

Now apply a voltage source ‘V’ to these plates, then all positive charges are accumulated on conductor similarly negative charges are accumulated on conductor  .

the charges give rise to a field   in between them called as induced electric field.

To find out capacitance, choose a co-ordinate system x, y and z as shown in the figure

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Noe the potential difference is .

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is directed from 2 to 1 work has to be done in opposite direction from 1 to 2.

Now . assume two conducting plates has equal surface area S

to find out at any point P in between the ‘2’ plates,  use the concept of infinite sheet of charge distributions with densities    and    . — from the positive charge distribution.

–with the negative charge distribution.

Electric field intensity at P is the sum of electric field intensities due to two infinite charge distributions

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Now the potential difference between the two conductors is .

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we know that .

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If the medium between two parallel plates is air  (or) free space (or) Vaccum  use   or else use .     (1 votes, average: 5.00 out of 5) Loading... ## Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.

Posted on Categories Electro Magnetic Theory