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QJEEM20Ph8d03: JEE Main 2020 Physics - Class 12 JEEM20Ph8d03 Solution: A parallel plate capacitor has plates of area A separated by…

JEE Main 2020 (Online) 7th January Morning Slot Physics - Capacitor Question 106 English A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + α\alpha x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression :

Multiple Choice Options

  • A A∈0Kd(1+(αd2)2){{A{ \in _0}K} \over d}\left( {1 + {{\left( {{{\alpha d} \over 2}} \right)}^2}} \right)
  • B A∈0Kd(1+αd2){{A{ \in _0}K} \over d}\left( {1 + {{\alpha d} \over 2}} \right)
  • C A∈0Kd(1+α2d22){{A{ \in _0}K} \over d}\left( {1 + {{{\alpha ^2}{d^2}} \over 2}} \right)
  • D A∈0Kd(1+αd){{A{ \in _0}K} \over d}\left( {1 + \alpha d} \right)

Step-by-Step Verified Solution

k(x) = K(1 + α\alpha x) JEE Main 2020 (Online) 7th January Morning Slot Physics - Capacitor Question 106 English Explanation
Capacitance of element dC1 = Kε0Adx{{K{\varepsilon _0}A} \over {dx}} = K(1+αx)ε0Adx{{K\left( {1 + \alpha x} \right){\varepsilon _0}A} \over {dx}}

1Ceq{1 \over {{C_{eq}}}} = ∫1dC1\int {{1 \over {d{C_1}}}}

= ∫0ddxKA(1+αx)ε0\int\limits_0^d {{{dx} \over {KA\left( {1 + \alpha x} \right){\varepsilon _0}}}}

= 1KAαε0ln⁡(1+αx){{1 \over {KA\alpha {\varepsilon _0}}}\ln \left( {1 + \alpha x} \right)}

Using expansion of ln (1 + x) keeping x ≪\ll 1

⇒\Rightarrow 1Ceq{1 \over {{C_{eq}}}} = 1KAαε0(αd−α2d22){{1 \over {KA\alpha {\varepsilon _0}}}\left( {\alpha d - {{{\alpha ^2}{d^2}} \over 2}} \right)}

⇒\Rightarrow Ceq = ε0KAd(1−αd2)−1{{{{\varepsilon _0}KA} \over d}{{\left( {1 - {{\alpha d} \over 2}} \right)}^{ - 1}}}

⇒\Rightarrow Ceq = ε0KAd(1+αd2){{{{\varepsilon _0}KA} \over d}\left( {1 + {{\alpha d} \over 2}} \right)}

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