JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 6)
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?
$${l \over 4}$$
$${l \over 2}$$
$${{2l} \over 3}$$
$${l \over 3}$$
Explanation
_5th_September_Evening_Slot_en_6_1.png)
Ci = $${{{\varepsilon _0}A} \over d} = {{{\varepsilon _0}lw} \over d}$$
Ui = $${1 \over 2}{C_i}{V^2}$$ = $${1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}$$
_5th_September_Evening_Slot_en_6_2.png)
Cf = C1 + C2
= $${{K{\varepsilon _0}{A_1}} \over d} + {{{\varepsilon _0}{A_2}} \over d}$$
= $${{K{\varepsilon _0}wx} \over d} + {{{\varepsilon _0}w\left( {l - x} \right)} \over d}$$
= $${{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]$$
$$ \therefore $$ Uf = $${1 \over 2}{C_f}{V^2}$$
= $${1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}$$
Given Uf = 2Ui
$$ \Rightarrow $$ $${1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}$$ = 2 $$ \times $$ $${1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}$$
$$ \Rightarrow $$ kx + l – x = 2l
$$ \Rightarrow $$ 4x – x = l
$$ \Rightarrow $$ 3x = l
$$ \Rightarrow $$ x = $${l \over 3}$$
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