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QJEEM20Ph388d: JEE Main 2020 Physics - Class 12 JEEM20Ph388d Solution: A parallel plate capacitor has plate of length 'l', width ‘w’…

A parallel plate capacitor has plate of length 'l', width ‘w’ and separation of plates is ‘d’. It is connected to a battery of emf V. A dielectric slab of the same thickness ‘d’ and of dielectric constant k = 4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

Multiple Choice Options

  • A l4{l \over 4}
  • B l2{l \over 2}
  • C 2l3{{2l} \over 3}
  • D l3{l \over 3}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 5th September Evening Slot Physics - Capacitor Question 95 English Explanation 1

Ci = ε0Ad=ε0lwd{{{\varepsilon _0}A} \over d} = {{{\varepsilon _0}lw} \over d}

Ui = 12CiV2{1 \over 2}{C_i}{V^2} = 12ε0lwdV2{1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2} JEE Main 2020 (Online) 5th September Evening Slot Physics - Capacitor Question 95 English Explanation 2

Cf = C1 + C2

= Kε0A1d+ε0A2d{{K{\varepsilon _0}{A_1}} \over d} + {{{\varepsilon _0}{A_2}} \over d}

= Kε0wxd+ε0w(l−x)d{{K{\varepsilon _0}wx} \over d} + {{{\varepsilon _0}w\left( {l - x} \right)} \over d}

= ε0wd[Kx+l−x]{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]

∴\therefore Uf = 12CfV2{1 \over 2}{C_f}{V^2}

= 12ε0wd[Kx+l−x]V2{1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}

Given Uf = 2Ui

⇒\Rightarrow 12ε0wd[Kx+l−x]V2{1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2} = 2 ×\times 12ε0lwdV2{1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}

⇒\Rightarrow kx + l – x = 2l

⇒\Rightarrow 4x – x = l

⇒\Rightarrow 3x = l

⇒\Rightarrow x = l3{l \over 3}

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