Two parallel plates have equal and opposite charges. When the space between them is evacuated,the electric field between the plates is $2 \times 10^5 \ V/m$. When the space is filled with a dielectric,the electric field becomes $1 \times 10^5 \ V/m$. Find the dielectric constant of the dielectric material.

$A$ parallel plate capacitor has a plate separation of $0.01\, mm$ and uses a dielectric (whose dielectric strength is $19\, kV/mm$) as an insulator. The maximum potential difference that can be applied to the terminals of the capacitor is......$V$

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 force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$. Half the gap between the two charges is filled by a dielectric slab of dielectric constant $K=4$. Then the new force of repulsion between those two charges becomes:

$A$ parallel plate capacitor has a capacitance of $C_0$. If a dielectric slab of relative permittivity $\varepsilon_r$ and thickness equal to one-fourth of the distance between the plates is inserted,the new capacitance is $C$. Then,the ratio $\frac{C}{C_0}$ is:

The capacitance of a parallel plate capacitor is $5\, \mu F$. When a glass slab of thickness equal to the separation between the plates is introduced between the plates,the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is