Physics, asked by bharati8348, 1 year ago

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Derive an expression for capacitance of parallel plate capacitor.

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Answers

Answered by ayush939194
2

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

begin mathsize 12px style E space equals space sigma over epsilon subscript 0 space end style .....................(1)

where ε0 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

begin mathsize 12px style V over d equals space fraction numerator begin display style bevelled Q over A end style over denominator epsilon subscript 0 end fraction space equals space fraction numerator Q over denominator A epsilon subscript 0 end fraction space........................... left parenthesis 2 right parenthesis

b u t space w e space h a v e space b y space d e f i n i t i o n space o f space c a p a c i tan c e comma space C space equals space Q over V

U sin g space e q n. left parenthesis 2 right parenthesis comma space w e space c a n space w w r i t e space C equals space Q over V equals space fraction numerator A epsilon subscript 0 over denominator d end fraction end style

----------------------------------

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


bharati8348: thanks
bharati8348: but where is diagram??
ayush939194: i can't send diagram because my net is too slow
bharati8348: :-(
Answered by Anonymous
0

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

begin mathsize 12px style E space equals space sigma over epsilon subscript 0 space end style .....................(1)

where ε0 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

begin mathsize 12px style V over d equals space fraction numerator begin display style bevelled Q over A end style over denominator epsilon subscript 0 end fraction space equals space fraction numerator Q over denominator A epsilon subscript 0 end fraction space........................... left parenthesis 2 right parenthesis

b u t space w e space h a v e space b y space d e f i n i t i o n space o f space c a p a c i tan c e comma space C space equals space Q over V

U sin g space e q n. left parenthesis 2 right parenthesis comma space w e space c a n space w w r i t e space C equals space Q over V equals space fraction numerator A epsilon subscript 0 over denominator d end fraction end style

----------------------------------

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

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