A capacitance of $2 \mathrm{\mu F}$ is required in an electrical circuit across a potential difference of $1.0 \mathrm{kV}$. Many $1 \mathrm{\mu F}$ capacitors are available which can withstand a potential difference of not more than $300 \mathrm{V}$. The minimum number of capacitors required to achieve this is

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: