A  CAPACITOR IS GIVEN WHICH HAS TWO dielectric mediums filled in it

the dielectric constant are k1 &k2  area = A

distn  of plates = d   the the distribution of mediums is symmetric about the

diagonal of the front face  i.e the mediums are seperated symmetry about diagonal

 

 well sorry,  that i have posted it in mechanics section

plz answer anybody

cmon yarrrrrrrrr!!!!!!!!!!!!!!!

  since the two types of dielectrics are seemed to be in series 1/dC = 1/dC''+ 1/dC'

i.e..   dC'' = k'' dx L / x tan                           dC' = k' dx L / (b-x)tan

you can get ot by intigrating the eq. 1/dC = 1/dC'' = 1/dC'

by the way L is the lenth of the plate and b is the width of the plate

Let us consider a strip of width dx at a distance x from the left end.

Here the thickness of the dielectric 1  =  xd /A .

Thickness of the dielectric 2  =  d( 1- x /A ). Then

P.D = _[0]^{[xd/A ]}  dz / e_0K1  + _{[xd/A ]}^[d]   dz / e_0K2 

          =  x d / e_{0K1 A} +   (  d - xd /A ) / e_0K2

         = x d / e_{0K1 A}  + d / e₀K₂  -  x d / e_{0K2 A}

          = d / e₀K₂ + ( x d / e_{0 A} ) ( 1/k1 -1/k2 )

So capacity of this small elementary capacitor under consideration is

dC =  x d b / [ d / e₀K₂ + ( x d / e_{0 A} ) ( 1/k1 -1/k2 ) ]

here b = breadth of each plate.

It is to be noted that now it is a strip so its length ( in the sense, length of a capacitor ) is b, breadth of each plate.

It is to be noted

dC = e₀ A dx / [ d  {  A /  k2 +  x(  k2 - k1 )  /  k1 k2  } ] 

       = e₀ k1 k2 A dx / [ d { A k1 + x( k2 - k1 ) } ]

This is the capacity of this strip or elementary capacitor.

Finaly

C = ( e₀ k1 k2 A / d ) _{[0 ]}^{[A ]} dx / [ A k1 + x ( k2 - k1 ) ]

   = ( e₀ k1 k2 A / d ) [ log A k2 - log Ak1]

  =  e₀ k1 k2 A  log ( k2 /k1)  / [ ( k2 -k1 ) d ]

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