Question 357: Electrostatics & Capacitance

Question

A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d < < a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is :

Accepted Answer

Let  the capacitance of the  upper  part of the strip of length dx is dC
1
and lower part of the strip is dC
2
$$ 	herefore $$   dC
1
= $${{{\varepsilon _0}\,adx} \over {d - y}}$$
and dC
2
= $${{K{\varepsilon _0}\,adx} \over y}$$
Here dC
1
and dC
2
are in series.
So, equivalent capacitance,
$${1 \over {dC}}$$  =  $${1 \over {d{C_1}}} + {1 \over {d{C_2}}}$$
$$ \Rightarrow $$    $${1 \over {dC}}$$  =  $${{d - y} \over {{\varepsilon _0}a\,dx}} + {y \over {{\varepsilon _0}K\,adx}}$$
$$ \Rightarrow $$   $${1 \over {dC}}$$  =  $${1 \over {{\varepsilon _0}\,adx}}$$ ( d $$-$$ y + $${y \over K}$$)
$$ \Rightarrow $$   dC = $${{{\varepsilon _0}\,adx} \over {\left( {d - y} ight) + {y \over K}}}$$
We can divide entire parallel plate capacitor into similar  part like the strip of length dx and all the strips will have common end  A  and B. So they are in parallel.
So equivalent capacitance is the sum of all the strips capacitance.
$$ 	herefore $$   $$\int {dC} $$ =  $$\int\limits_0^a {{{{\varepsilon _0}\,adx} \over {\left( {d - y} ight) + {y \over K}}}} $$
$$ \Rightarrow $$   C = $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {\left( {d - y} ight) + {y \over K}}}} $$
From above diagram you can find this $$ 	o $$
tan$$	heta $$  =  $${y \over x}$$
Also tan$$	heta $$  =  $${d \over a}$$
$$ 	herefore $$   $${y \over x}$$  =  $${d \over a}$$
$$ \Rightarrow $$   y  =  $${d \over a}$$ x
By putting this value of y in the integration we get,
C  =  $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d - {d \over a}x + {d \over {Ka}}x}}} $$
=  $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d + \left( {{1 \over K} - 1} ight){d \over a}x}}} $$
=  $$\int\limits_0^a {{{{\varepsilon _0}\,{a^2}\,dx} \over {da + \left( {{{1 - K} \over K}} ight)xd}}} $$
=  $$\int\limits_0^a {{{K{\varepsilon _0}\,{a^2}\,dx} \over {Kda + \left( {1 - K} ight)xd}}} $$
=  $${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} ight)}}\left[ {\ln \left( {\left( {1 - K} ight)x + Ka} ight)} ight]_0^a$$
=  $${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} ight)}}\left[ {\ln \left( a ight) - \ln \left( {Ka} ight)} ight]$$
=  $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} ight)}}\ln \left( {{a \over {Ka}}} ight)$$
=  $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} ight)}}\ln \left( {{1 \over K}} ight)$$
=  $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} ight)}}\ln \left( K ight)$$

Answer

$${{K{ \in _0}{a^2}} \over {2d(K + 1)}}$$

Answer

$${{K{ \in _0}{a^2}} \over {d(K - 1)}}\ln K$$

Answer

$${{K{ \in _0}{a^2}} \over d}\ln K$$

Answer

$${1 \over 2}{{K{ \in _0}{a^2}} \over d}$$

Electrostatics & Capacitance

A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d < < a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is :

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