Capacitance of Parallel Plate Capacitor

Advertisements

Advertisements3

Advertisements4

Advertisements6

Advertisements66

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 $\epsilon$ .

Note: There is a rating embedded within this post, please visit this post to rate it.

Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems. View all posts by Lakshmi Prasanna Ponnala

Capacitance of Parallel Plate Capacitor

Like this:

Related

Author: Lakshmi Prasanna Ponnala

Like this:

Related

Author: Lakshmi Prasanna Ponnala

Discover more from ece4uplp