A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E _{1}$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _{2}$. The ratio $E _{1} / E _{2}$ will be :
$\frac{1}{10}$
$\frac{2}{5}$
$\frac{5}{13}$
$\frac{5}{26}$
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
Explain the effect of dielectric on capacitance of parallel plate capacitor and obtain the formula of dielectric constant.
A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by $Q_0, V_0, E_0$ and $U_0$ respectively. A dielectric slab is introduced between plates of capacitor but battery is still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ related to previous ones are
A parallel plate capacitor, partially filled with a dielectric slab of dielectric constant $K$ , is connected with a cell of emf $V\ volt$ , as shown in the figure. Separation between the plates is $D$ . Then
What are polar and non-polar molecules ?