Capacitance of A parralel plate capaciter

### Asked by Poojakaushik303 | 26th Nov, 2019, 11:18: AM

Expert Answer:

###
###
Let us assume two parallel plates of area A are kept so that their separation distance is d.
One plate is positively charged and the other is negatively charged.
This charging can be done by connecting the parallel plates across the terminals of battery.
Let σ be the surface charge density. Compare to the dimensions of plate, the separation distance is very small.
With this assumption, we can have electrical field due the charged plate is same as that of infinitely charged plate.
Since the charges on the plate are opposite, electric fields due to each charged plate is in same direction inside
the region between plates as shown in figure.
Hence the net electric field E in the inside region between plates is given by
.....................(1)
where ε_{0} is the permitivity of free space. But we have E = V/d , where V is potential difference between plates.
Also σ = Q/A , where Q is the magnatiude of total charge on each plates.
Using these relations Eqn.(1) is rewritten as
----------------------------------
Capacitance of capacitor is increaed by having dielectric materials between parallel plates.
If K is the dielectric constant of dielectric material present betwenn parallel plates,
then the capacitance of capacitor becomes K times of capacitance of paralle plate capacitor without dieletric material

###
Let us assume two parallel plates of area A are kept so that their separation distance is d.
One plate is positively charged and the other is negatively charged.
This charging can be done by connecting the parallel plates across the terminals of battery.
Let σ be the surface charge density. Compare to the dimensions of plate, the separation distance is very small.
With this assumption, we can have electrical field due the charged plate is same as that of infinitely charged plate.
Since the charges on the plate are opposite, electric fields due to each charged plate is in same direction inside
the region between plates as shown in figure.
Hence the net electric field E in the inside region between plates is given by
.....................(1)
where ε_{0} is the permitivity of free space. But we have E = V/d , where V is potential difference between plates.
Also σ = Q/A , where Q is the magnatiude of total charge on each plates.
Using these relations Eqn.(1) is rewritten as
----------------------------------
Capacitance of capacitor is increaed by having dielectric materials between parallel plates.
If K is the dielectric constant of dielectric material present betwenn parallel plates,
then the capacitance of capacitor becomes K times of capacitance of paralle plate capacitor without dieletric material

_{0}is the permitivity of free space. But we have E = V/d , where V is potential difference between plates.

### Answered by Shiwani Sawant | 26th Nov, 2019, 02:33: PM

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