A parallel plate capacitor having plates of area S and plate separation d, has capacitance `C_1` in air. When two dielectrics of different relative primitivities (`epsilon_1=2` and `epsilon_2=4`) are introduced between the two plates as shown in the figure, the capacitance becomes `C_2`. The ratio `C_2/C_1` is

A parallel plate capacitor having plates of area S and plate separation d, has capacitance C_1 in air. When two dielectrics of different relative primitivities ( epsilon_1=2epsilon_0 and epsilon_2=4epsilon_0 ) are introduced between the two plates as shown in the figure, the capacitance becomes C_2 . The ratio C_2/C_1 is

For a parallel plate capacitor of capacitance C and plate separation d,

A parallel plate capacitor, with plate area A and plate separation d, is filled with a dielectric slabe as shown. What is the capacitance of the arrangement ?

Three dielectrics of relative permittivities epsilon_(r_(1))=1,epsilon_(r_(2))=2 , and epsilon_(r_(3))=3 are introduced in a parallel plate capacitor of plate A and B. .

A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be :

A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be :