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Correct option is B)

Initial capacitance$=dϵ_{∘}A $

When it is half filled by a dielectric of dielectric constant $k$ then

$C_{1}=2d kϵ_{∘}A =2kdϵ_{∘}A $

and $C_{2}=2d ϵ_{∘}A =d2ϵ_{∘}A $

$C1 =C_{1}1 +C_{2}1 $

$=2ϵ_{∘}Ad (k1 +1)$

$=2ϵ_{∘}Ad (51 +1)$

$=2ϵ_{∘}Ad (56 )$

$=5ϵ_{∘}A3d $

$∴C=3d5ϵ_{∘}A $

Hence, $%$ increase in capacitance is

$=⎝⎜⎜⎛ dϵ_{∘}A d5ϵ_{∘}A −dϵ_{∘}A ⎠⎟⎟⎞ ×100=32 ×100=66.6%$

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