The electrostatic force between the metal plates of an isolated parallel plate capacitor $C$ having a charge $Q$ and area $A$, is 

The figure shows an experimental plot discharging of a capacitor in an $RC$ circuit. The time constant $\tau$ of this circuit lies between : 

Two capacitors $C_1$ and $C_2$ are charged to $120 \mathrm{V}$ and $200 \mathrm{V}$, respectively. It is found that by connecting them together, the potential on each one can be made zero. Then 

In the given circuit, charge $Q_2$ on the $2\mu F$ capacitor changes as $C$ is varied from $1\mu F$ to $3\mu F.$ $Q_2$ as a function of ${{}^{\text{'}}C}^{\text{'}}$ is given properly by: (figures are drawn symmetrically and are not to scale)

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge $Q$ (having a charge equal to the sum of the charges on the $4\mu F$ and $9\mu F$ capacitors), at a point distance $30 \mathrm{m}$ from it, would equal :

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: