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


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

C'=ε0Ad-t1-1KC'=ε0A3d-2d1-1; for conductor K=C'=ε0A3d-2dC'=ε0Ad=3ε0A3dC'=3C0.


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