# 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 $M_{2}$ similarly negative charges are accumulated on conductor  $M_{1}$ .

$\therefore$ the charges give rise to a field $\overrightarrow{E}$  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

$\therefore&space;V&space;=-\int&space;\overrightarrow{E}.\overrightarrow{dl}$.

Noe the potential difference is $V=V_{2}-V_{1}$.

$V_{12}=V_{2}-V_{1}$.

$\therefore&space;V&space;=-\int_{1}^{2}&space;\overrightarrow{E_{induced}}.\overrightarrow{dl}$.

$\overrightarrow{E}$  is directed from 2 to 1 work has to be done in opposite direction from 1 to 2.

Now $\rho&space;_{s}=\frac{Q}{S}$ . assume two conducting plates has equal surface area S

$\therefore$  to find out $\overrightarrow{E}$ at any point P in between the ‘2’ plates,  use the concept of infinite sheet of charge distributions with densities  $\rho&space;_{s}$  and   $-\rho&space;_{s}$ .

$E_{\rho&space;_{s}}&space;=&space;\frac{\rho&space;_{s}}{2\epsilon&space;_{o}}(-\overrightarrow{a_{z}})$ — from the positive charge distribution.

$E_{-\rho&space;_{s}}&space;=&space;\frac{-\rho&space;_{s}}{2\epsilon&space;_{o}}(\overrightarrow{a_{z}})$ –with the negative charge distribution.

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

$\overrightarrow{E_{total}}&space;=&space;\overrightarrow{E_{\rho&space;_{s}}}+\overrightarrow{E_{-\rho&space;_{s}}}$ .

$\overrightarrow{E_{total}}&space;=&space;\frac{\rho&space;_{s}}{2\epsilon&space;_{o}}(-\overrightarrow{a_{z}})+&space;\frac{-\rho&space;_{s}}{2\epsilon&space;_{o}}(\overrightarrow{a_{z}})$ .

$\overrightarrow{E_{total}&space;}=&space;\frac{\rho&space;_{s}}{\epsilon&space;_{o}}(-\overrightarrow{a_{z}})$ .

Now the potential difference between the two conductors is $\therefore&space;V&space;=-\int_{1}^{2}&space;\overrightarrow{E_{induced}}.\overrightarrow{dl}$ .

$\therefore&space;V&space;=-\int_{1}^{2}&space;\frac{\rho&space;_{s}}{\epsilon&space;_{o}}(-\overrightarrow{a_{z}}).\overrightarrow{dl}$ .

we know that $\overrightarrow{dl}&space;=&space;dx\&space;\overrightarrow{a_{x}}+&space;dy\&space;\overrightarrow{a_{y}}+&space;dz\&space;\overrightarrow{a_{z}}$ .

$\therefore&space;V&space;=\int_{1}^{2}&space;\frac{\rho&space;_{s}}{\epsilon&space;_{o}}(\overrightarrow{a_{z}}).\&space;dz&space;\&space;\overrightarrow{a_{z}}$ .

$V&space;=\int_{0}^{d}&space;\frac{\rho&space;_{s}}{\epsilon&space;_{o}}.\&space;dz$ .

$V&space;=&space;\frac{\rho&space;_{s}d&space;}{\epsilon&space;_{o}}$ .

$V&space;=&space;\frac{Q\&space;d&space;}{S\&space;\epsilon&space;_{o}}&space;\&space;\because&space;\rho&space;_{s}=\frac{Q}{S}$ .

$\therefore&space;C=&space;\frac{Q}{V}=\frac{S\&space;\epsilon&space;_{o}&space;}{d}$ .

If the medium between two parallel plates is air  (or) free space (or) Vaccum  use $\epsilon&space;_{o}$  or else use $\epsilon$ .

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