Show that if a dielectric of thickness t and with the same area as the plates of a parallel plate capacitor is introduced, the capacitance of the capacitor is given by
`C=(epsi_(0)A)/((d-t+t/epsi_(r))`

Show that if a dielectric of thickness t and with the same area as the plates of a parallel plate capacitor is introduce the capacitance of the capacitor is given by C=(epsi_(0)A)/((d-t+t/epsi_(r))

A dielectric slab is placed between the plates of a parallel plate capacitor. Its capacitance

If a thin metal foil of same area is placed between the two plates of a parallel plate capacitor of capacitance C, then new capacitance will be

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor, but has a thickness 3d//4 . Find the ratio of the capacitanace with dielectric inside it to its capacitance without the dielectric.

A copper plate of thickness b is placed inside a parallel plate capacitor of plate distance d and area A as shown in figure. The capacitance of capacitor is

If a dielectric slab of thickness 5 mm and dielectric constant K=6 is introduced between the plates of a parallel plate air capacitor, with plate separation of 8 mm, then its capacitance is