Effective capacitance of parallel combination of two capacitors C₁ and C₂ is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C₂ is 4 times that of C₁. If these capacitors are connected in series, their effective capacitance will be :

A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + $$\alpha $$x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression :

In the given circuit, the charge on 4 $$\mu $$F capacitor will be :

Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K₁, K₂ and K₃. The first capacitor is filled as shown in fig.I, and the second one is filled as shown in fig II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E₁ refers to capacitor (I) and E₂ to capacitor (II)):