A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is(A) \(C=\cfrac{Aε_0}{d}\left(\cfrac{k+3}{4k}\right)\)(B) \(C=\cfrac{Aε_0}{d}\left(\cfrac{2k}{k+3}\right)\)(C) \(C=\cfrac{Aε_0}{d}\left(\cfrac{k+3}{2k}\right)\)(D) \(C=\cfrac{Aε_0}{d}\left(\cfrac{4k}{k+3}\right)\)

A slab of material of dielectric constant k has the same area A as the

1.	A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is(A) \(C=\cfrac{Aε_0}{d}\left(\cfrac{k+3}{4k}\right)\)(B) \(C=\cfrac{Aε_0}{d}\left(\cfrac{2k}{k+3}\right)\)(C) \(C=\cfrac{Aε_0}{d}\left(\cfrac{k+3}{2k}\right)\)(D) \(C=\cfrac{Aε_0}{d}\left(\cfrac{4k}{k+3}\right)\)
Answer» Correct option is: (D) \(C=\cfrac{Aε_0}{d}\left(\cfrac{4k}{k+3}\right)\)

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