A capacitor has capacitance $5 \mu F$ when it's parallel plates are separated by air medium of thickness $d$. A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\frac{ d }{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be $..........\mu F$

How does the polarised dielectric modify the original external field inside it ?

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is introduced between the plates which results in

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 share the charge and the common potential is $V'$. The dielectric constant $K$ is

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant $K $ is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.

A parallel plate capacitor of plate area $A$ and plate seperation $d$ is charged to potential difference $V$ and then the battery is disconnected. Aslab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q, E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then