If a slab of insulating material $4 \times 10^{-3} \ m$ thick is introduced between the plates of a parallel plate capacitor,the separation between the plates has to be increased by $3.5 \times 10^{-3} \ m$ to restore the capacity to its original value. The dielectric constant of the material will be:

$A$ parallel plate capacitor has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?

$A$ slab of dielectric constant $K$ has the same cross-sectional 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 when the slab is inserted between the plates will be (Given $C_{0} =$ capacitance of capacitor with air as medium between plates):

$A$ capacitor with air as the dielectric is charged to a potential of $100\;V$. If the space between the plates is now filled with a dielectric of dielectric constant $K = 10$,the potential difference between the plates will be: (in $;V$)

$A$ parallel-plate capacitor with plate area $A$ has a separation $d$ between the plates. Two dielectric slabs of dielectric constants $K_{1}$ and $K_{2}$,each having an area of $A/2$ and thickness $d/2$,are inserted in the space between the plates as shown in the figure. The equivalent capacitance of the capacitor will be:

When a dielectric material is introduced between the plates of a charged condenser,what happens to the electric field between the plates?