A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d < < a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is :

In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C₁ as a function of time will be given by $$\left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$$

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :

A capacitor C₁ = 1.0 $$\mu $$F is charged up to a voltage V = 60 V by connecting it to battery B through switch (1). Now C₁ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $${C_2} = 3.0\mu F$$ and C₃ = 6.0 $$\mu $$F through switch (2), as shown in the figure. The sum of final charges on C₂ and C₃ is :