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 V =-\int \overrightarrow{E}.\overrightarrow{dl}.

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

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

\therefore V =-\int_{1}^{2} \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 _{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 _{s}  and   -\rho _{s} .

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

E_{-\rho _{s}} = \frac{-\rho _{s}}{2\epsilon _{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}} = \overrightarrow{E_{\rho _{s}}}+\overrightarrow{E_{-\rho _{s}}} .

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

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

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

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

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

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

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

V = \frac{\rho _{s}d }{\epsilon _{o}} .

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

\therefore C= \frac{Q}{V}=\frac{S\ \epsilon _{o} }{d} .

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

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Author: vikramarka

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