Between the plates of a parallel plate condenser there is $1\,mm$ thick paper of dielectric constant $4$. It is charged at $100\;volt$. The electric field in $volt/metre$ between the plates of the capacitor is

The gap between the plates of a parallel plate capacitor of area $A$ and distance between plates $d$, is filled with a dielectric whose permittivity varies linearly from ${ \varepsilon _1}$ at one plate to ${ \varepsilon _2}$ at the other. The capacitance of capacitor is

Figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant $3$. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric

The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ - 12}}farads$. The dielectric constant of wax is

Two parallel plate capacitors of capacity $C$ and $3\,C$ are connected in parallel combination and charged to a potential difference $18\,V$. The battery is then disconnected and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $9$. The final potential difference across the combination of capacitors will be $V$

A parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K =4$. The thickness of the dielectric material is $x$, where $x < d$.

Let $C_1$ and $C_2$ be the capacitance of the system for $x =\frac{1}{3} d$ and $x =\frac{2 d }{3}$, respectively. If $C _1=2 \mu F$ the value of $C _2$ is $........... \mu F$