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two conductors of thickness d are inserted inside a parallel capacitor of thickness 3d and capacitance C0 . The capacitance of new arrangement.

answer fast please

Dear Student,

Please find below the solution to the asked query:

Let the plate area of the capacitor is *A*, given that the separation between the plates is *3d*. Then according to the capacitance of the parallel plate capacitor is,

${C}_{0}={\frac{{\epsilon}_{0}A}{3d}}_{}$

If a conducting slab is introduced between the plates of the capacitor, the effective capacitance is,

$C\text{'}=\frac{{\epsilon}_{0}A}{d-t\left(1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$K$}\right.}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow C\text{'}=\frac{{\epsilon}_{0}A}{3d-2d\left(1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\infty $}\right.}\right)};\left(forconductorK=\infty \right)\phantom{\rule{0ex}{0ex}}\Rightarrow C\text{'}=\frac{{\epsilon}_{0}A}{3d-2d}\phantom{\rule{0ex}{0ex}}\Rightarrow C\text{'}=\frac{{\epsilon}_{0}A}{d}=3\left(\frac{{\epsilon}_{0}A}{3d}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow C\text{'}=3{C}_{0}$.

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