Capacitance

In the given circuit diagram, $E=5V,r=1\Omega\,,R_{2}=4\Omega\,,R_{1}=R_{3}=1\Omega\,$ and $C=3\mu\,F$ Then, what will be the numerical value of charge on each plates of the capacitor.

The equivalent capacitance of the combination shown in the figure is:

The capacitance of a parallel plate capacitor with air as medium is $6\mu\,\text{F}$. With the introduction of a dielectric medium, the capacitance becomes $30\mu\,\text{F}$. The permittivity of the medium is:…

A parallel-plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant $K_{1}$ and $K_{2}$ of same area $\frac{A}{2}$and thickness $\frac{d}{2}$are inserted in…

In the circuit shown, charge on the$5\mu\,F$ capacitor is :

A slab of dielectric constant $K$ has the same crosssectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$, where $d$ is the separation of the plates. The capacitance of the capacitor…

A capacitor of capacitance $C=1\mu\,F$ is suddenly connected to a battery of $100V$through a resistance $R=100\Omega\,.$ The time taken for the capacitor to be charged to get $50V$ is: (Take$\ln\,2=0.69$)

The material filled between the plates of a parallel plate capacitor has resistivity $200\Omega\,m$. The value of capacitance of the capacitor is $2\mathrm{pF}$. If a potential difference of $40V$ is applied across the…

A parallel plate capacitor has plates of area A separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $K\left(x\right)=K_{0}\left(1+\alpha\,x\right)$ where…

The charge on capacitor of capacitance $15\mu\,F$ in the figure given below is

The distance between two plates of a capacitor is $d$ and its capacitance is $C_{1}$, when air is the medium between the plates. If a metal sheet of thickness $\frac{2d}{3}$ and of the same area as plate is introduced…

A parallel plate capacitor of capacitance $12.5 \mathrm{pF}$ is charged by a battery connected between its plates to potential difference of $12.0 \mathrm{~V}$. The battery is now disconnected and a dielectric slab…