In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance $C$ will be :

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

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1,{ K}_2,$ $K_3,{ K}_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be-

A parallel plate capacitor is of area $6{ \mathrm{cm}}^2$ and a separation $3 \mathrm{mm}.$ The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1=10,{ K}_2=12$ and $K_3=14.$ The dielectric constant of a material which when fully inserted in the above capacitor, gives the same capacitance would be:

In the figure shown below, the charge on the left plate of the $10 \mathrm{\mu F}$ capacitor is $-30 \mathrm{\mu C}.$ The charge on the right plate of the $6 \mathrm{\mu F}$ capacitor is:

In the figure shown, after the switch $S$ is turned position $A$ to position $B,$ the energy dissipated in the circuit in terms of capacitance $C$ and total charge $Q$ is: