The area of the plates of a parallel plate capacitor is $A$ and the gap between them is $d$. The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance $‘y’$ from one plate as : $K = \lambda \ sec(\pi y/2d)$, where $\lambda $ is a dimensionless constant. The capacitance of this capacitor is
$\pi \varepsilon_0\lambda A / 2d$
$\pi \varepsilon_0\lambda A /d$
$2\pi \varepsilon_0\lambda A /d$
none
Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be
With the rise in temperature, the dielectric constant $K$ of a liquid
Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes $6\,times$.
Reason : Capacity of the capacitor does not depend upon the nature of the material.
Write the capacitance of parallel plate capacitor with medium of dielectric of dielectric constant $\mathrm{K}$.
The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
$K(x) = K_0 + \lambda x$ ( $\lambda =$ constant)
The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation