An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air cored parallel capacitor charged to a potential $V$. The two capacitors share charges and the common potential is $V$. The dielectric constant $K$ is
$\frac{V^{\prime}-V}{V^{\prime}+V}$
$\frac{V^{\prime}-V}{V^{\prime}}$
$\frac{V^{\prime}-V}{V}$
$\frac{V-V^{\prime}}{V^{\prime}}$
Which one statement is correct ? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are ${Q_o},\;{V_o},\;{E_o}$ and ${U_o}$ respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values $Q,\;V,\;E$ and $U$ are in relation with the initially stated as
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
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 $\mathrm{k}(\mathrm{x})=\mathrm{K}(1+\alpha \mathrm{x})$ where $\mathrm{x}$ is the distance measured from one of the plates. If $(\alpha \text {d)}<<1,$ the total capacitance of the system is best given by the expression
Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the dielectric constant of a material, which when fully filled between the plates produces same capacitance?
In a parallel plate capacitor the separation between the plates is $3\,mm$ with air between them. Now a $1\,mm$ thick layer of a material of dielectric constant $2$ is introduced between the plates due to which the capacity increases. In order to bring its capacity to the original value the separation between the plates must be made......$mm$