Capacitors

Find equivalent capacitance between $X$ and $Y$ if each capacitor is $4 \mathrm{\mu F}$.

An uncharged capacitor with a solid dielectric is connected to a similar air capacitor charged to a potential of $V_0.$ If the common potential after sharing of charges becomes $V\mathit{,}$ then the dielectric constant of the dielectric must be

Two parallel plate capacitors of capacitances $C$ and $2C$ are connected in parallel and charged to a potential difference $V.$ The battery is then disconnected and the region between the plates of the capacitor $C$ is completely filled with a material of dielectric constant $K\mathit{.}$ The potential difference across the capacitors now becomes -

A network of resistances, cell and capacitor $C (=2 \mathrm{\mu F})$ is shown in adjoining figure. In steady state condition, the charge on $2 \mathrm{\mu F}$ capacitor is $Q$ while $R$ is unknown resistance. Values of $Q$ and $R$ are, respectively,

In the given circuit, with steady current, the potential drop across the capacitor must be,

A capacitor of $1 \mu \mathrm{F}$ withstands a maximum voltage of $6 \mathrm{kV}$ while another capacitor of $2 \mathrm{\mu F}$ withstands a maximum voltage of $4 \mathrm{kV}$. If the two capacitors are connected in series, the system will withstand a maximum voltage of,

A parallel plate capacitor with air between the plates is charged to potential difference of $500 \mathrm{V}$ and then insulated. A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes $75 \mathrm{V}$. The dielectric constant of plastic is,

The capacitance of a parallel plate capacitor with air as dielectric is $C$. If a slab of dielectric constant $K$ and of the same thickness as the separation between the plates is introduced so as to fill ${\frac{1}{4}}^{\mathrm{th}}$ of the capacitor (shown in figure), then the new capacitance is,

The plates of a parallel plate capacitor are charged up to $100 \mathrm{volts}$. A $2 \mathrm{mm}$ thick dielectric slab is inserted between the plates. Then, to maintain same potential difference, the distance between the capacitor plates is increased by $1.6 \mathrm{mm}$. The dielectric constant of the plate is,

Two capacitors of capacitances $1 \mathrm{\mu F}$ and $\mathrm{C} \mathrm{\mu F}$ are connected in series and the combination is charged to a potential difference of $120 \mathrm{V}$. If the charge on the combination is $80 \mathrm{\mu C}$, the energy stored in the capacitor of capacity $\mathrm{C}$ in $\mathrm{\mu J}$ is,

The capacitance of a parallel plate capacitor with air as the medium is $3 \text{μF}.$ As a dielectric is introduced between the plates, the capacitance becomes $15 \text{μF}\mathit{.}$ The permittivity of the medium in ${\mathrm{C}}^2 {\mathrm{N}}^{-1} {\mathrm{m}}^{-2}$ is,

Identical dielectric slabs are inserted into two identical capacitors $A$ and $B.$ These capacitors and a battery are connected as shown in the figure. Now, the slab of capacitor $B$ is pulled out with battery remaining connected.

Two identical capacitors have the same capacitance $C.$ One of them is charged to potential $V_1$ and the other to $V_2.$ The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is,

The plate of a parallel plate capacitor are separated by $d \mathrm{cm}.$ A plate of thickness $t \mathrm{cm}$ with dielectric constant $k_1$ is inserted and the remaining space is filled with a plate of dielectric constant $k_2.$ If $Q$ is the charge on the capacitor and area of plates is $A {\mathrm{cm}}^2$ each, then the potential difference between the plates is,

Two capacitors of capacities $C$ and $2C$ are connected in parallel and then connected in series with a third capacitor of capacity $3C\mathit{.}$ The combination is charged with $V$ volt. The charge on capacitor of capacity $C$ is,

The capacity of a capacitor is $4\times {10}^{-6} \mathrm{F}$ and its potential is $100 \mathrm{V}$. The energy released on discharging it fully will be

During charging a capacitor, variations of potential $V$ of the capacitor with time $t$ is shown as,

In the given circuit, $C_1=C_2=C_3=C$ initially. Now, a dielectric slab of dielectric constant $K=\frac{3}{2}$ is inserted in $C_2.$

The equivalent capacitance become