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Correct option is C)

Refer image 1.

Let $k(x)=k(1+αx)$be the dielectric constant

The capacitor can be considered to be a series combination of a large number of capacitors of thickness $Δx$.

Refer image 2.

$C_{eq}1 =∑k(x)Aϵ_{0}Δx =∫_{0}k(x)Aϵ_{0}dx $

where $Δx→0$

$C_{eq}1 =∫_{0}k(1+αx)Aϵ_{0}dx =KAϵ_{0}1 ∫_{0}1+αxdx $

$=KAϵ_{0}1 g_{e}(11+αd )×α1 $

$=KαAϵ_{0}1 g_{e}(1+αd)$

Now, $g(1+x)=x−2x_{2} +3x_{3} −−−−−−−$

$∴g_{e}(1+αd)≃αd−2α_{2}d_{2} $

$∴C_{eq}1 =kαAϵ_{e}1 g_{e}(1+αd)$

$=kαAϵ_{0}1 [αd−2α_{2}d_{2} +−−−−−−−]$

$≃kαAϵ_{0}αd [1−2αd ]≃KAϵ_{0}d [1−2αd ]$

$C_{eq}=d[1−2αd ]KAϵ_{0} =dKAϵ_{0} [1−2αd ]_{−1}$

Now, $(1−x)_{−1}≃1+x$ $[∵x<<1]$

$∴C_{eq}=dKAϵ_{0} [1+2αd ]$

option (C) is correct.

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