In the figure,a capacitor is filled with dielectrics. The resultant capacitance is

Consider a parallel plate capacitor with plates in the shape of a square in the $XY$-plane. The gap between the plates is filled with a dielectric material. The dielectric constant $k$ varies along the $X$-axis as $k(x) = \left[1 + \left(\frac{x}{L}\right)^\alpha\right]$,where $\alpha$ is a constant. Let $C_d$ and $C_a$ be the capacitance in the presence of the dielectric and air,respectively. If the ratio $\frac{C_d}{C_a} = \frac{7}{6}$,then the value of $\alpha$ must be

The plates of a parallel plate capacitor are charged up to $100\,V$. $A$ $2\,mm$ thick dielectric plate is inserted between the plates. To maintain the same potential difference,the distance between the capacitor plates is increased by $1.6\,mm$. The dielectric constant of the plate is:

Initially,the circuit is in a steady state. Now,one of the capacitors is filled with a dielectric of dielectric constant $K = 2$. Find the heat loss in the circuit due to the insertion of the dielectric.

Two metal plates each of area $A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $d$. $A$ metal plate of thickness $\frac{d}{2}$ and of same area $A$ is inserted between the plates to form two capacitors of capacitances $C_1$ and $C_2$ as shown in the figure. If the effective capacitance of the two capacitors is $C^{\prime}$ and the capacitance of the capacitor initially is $C$,then $\frac{C^{\prime}}{C}$ is

$A$ parallel plate air capacitor has capacitance $C_p$. It is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$. Now its capacity becomes $C_K$. The ratio $C_P$ to $C_K$ is