A dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$. The induced charge $q^{\prime}$ on the surface of slab is given by
$q^{\prime}=q-\frac{q}{K}$
$q^{\prime}=-q+\frac{q}{K}$
$q^{\prime}=q\left[\frac{1}{K}+1\right]$
$q^{\prime}=-q\left(1+\frac{1}{K}\right)$
Explain the difference in the behaviour of a conductor and dielectric in the presence of external electric field.
Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf, $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$, then the amount of charge which will flow through the battery is (neglect internal resistance of the battery)
A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K (=4) .$ The ratio of new capacitance to its original capacitance will be,
Write the relation between $\vec P$ and $\vec E$ for a linear isotropic dielectric.
A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$