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……$mm$

A parallel plate capacitor is formed by two plates each of area $30 \pi\, cm ^{2}$ separated by $1\, mm$. A material of dielectric strength $3.6 \times 10^{7} \,Vm ^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6}\, C$, the value of dielectric constant of the material is

$\left\{ Use : \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right\}$

Following operations can be performed on a capacitor : $X$ – connect the capacitor to a battery of $emf$ $E.$ $Y$ – disconnect the battery $Z$ – reconnect the battery with polarity reversed. $W$ – insert a dielectric slab in the capacitor

There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant $K$ while the other one is charged to potential $V$ having air between its plates. If two capacitors are joined end to end, the common potential will be

A parallel plate capacitor Air filled with a dielectric whose dielectric constant varies with applied voltage as $K = V$. An identical capacitor $B$ of capacitance $C_0$ with air as dielectric is connected to voltage source $V_0 = 30\,V$ and then connected to the first capacitor after disconnecting the voltage source. The charge and voltage on capacitor.