JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 1)
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 :
_9th_January_Morning_Slot_en_1_1.png)
$${{K{ \in _0}{a^2}} \over {2d(K + 1)}}$$
$${{K{ \in _0}{a^2}} \over {d(K - 1)}}\ln K$$
$${{K{ \in _0}{a^2}} \over d}\ln K$$
$${1 \over 2}{{K{ \in _0}{a^2}} \over d}$$
Explanation
_9th_January_Morning_Slot_en_1_2.png)
_9th_January_Morning_Slot_en_1_3.png)
Let the capacitance of the upper part of the strip of length dx is dC1 and lower part of the strip is dC2
$$ \therefore $$ dC1 = $${{{\varepsilon _0}\,adx} \over {d - y}}$$
and dC2 = $${{K{\varepsilon _0}\,adx} \over y}$$
Here dC1 and dC2 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 = $${{{\varepsilon _0}\,adx} \over {\left( {d - y} \right) + {y \over K}}}$$
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.
$$ \therefore $$ $$\int {dC} $$ = $$\int\limits_0^a {{{{\varepsilon _0}\,adx} \over {\left( {d - y} \right) + {y \over K}}}} $$
$$ \Rightarrow $$ C = $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {\left( {d - y} \right) + {y \over K}}}} $$
From above diagram you can find this $$ \to $$
_9th_January_Morning_Slot_en_1_4.png)
tan$$\theta $$ = $${y \over x}$$
Also tan$$\theta $$ = $${d \over a}$$
$$ \therefore $$ $${y \over x}$$ = $${d \over a}$$
$$ \Rightarrow $$ y = $${d \over a}$$ x
By putting this value of y in the integration we get,
C = $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d - {d \over a}x + {d \over {Ka}}x}}} $$
= $$\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d + \left( {{1 \over K} - 1} \right){d \over a}x}}} $$
= $$\int\limits_0^a {{{{\varepsilon _0}\,{a^2}\,dx} \over {da + \left( {{{1 - K} \over K}} \right)xd}}} $$
= $$\int\limits_0^a {{{K{\varepsilon _0}\,{a^2}\,dx} \over {Kda + \left( {1 - K} \right)xd}}} $$
= $${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} \right)}}\left[ {\ln \left( {\left( {1 - K} \right)x + Ka} \right)} \right]_0^a$$
= $${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} \right)}}\left[ {\ln \left( a \right) - \ln \left( {Ka} \right)} \right]$$
= $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( {{a \over {Ka}}} \right)$$
= $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( {{1 \over K}} \right)$$
= $${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( K \right)$$
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