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,

$$\begin{gathered} C\text{'}=\frac{{\epsilon }_{0}A}{d-t\left(1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$K$}\right.}\right)} \\ \Rightarrow C\text{'}=\frac{{\epsilon }_{0}A}{3d-2d\left(1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\infty $}\right.}\right)}; \left(for conductor K=\infty \right) \\ \Rightarrow C\text{'}=\frac{{\epsilon }_{0}A}{3d-2d} \\ \Rightarrow C\text{'}=\frac{{\epsilon }_{0}A}{d}=3\left(\frac{{\epsilon }_{0}A}{3d}\right) \\ \Rightarrow C\text{'}=3C_0 \end{gathered}$$.

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