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I1 : A slab of material of dielectric constant K has same area as the plates of a parallel plateactor but has thickness 34(d), where d is the separation of plates. How is the capacitancechanged, when the slab is inserted between the plates?

Answer»

C =eo *A/(d-t + t/k)

C =eo *A/ (d- 3d/4 + 3d/4k)

C = (4k/k+3) (eo *A /d)

C = (4k/k+3) * Co



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