When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

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

Capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$, dielectric constant of paper is $2.5$ and width of paper is $400\,mm$, then length of foil will be…..$m$

A parallel plate capacitor is to be designed, using a dielectric of dielectric constant $5$, so as to have a dielectric strength of $10^9\;Vm^{-1}$ . If the voltage rating of the capacitor is $12\;kV$, the minimum area of each plate required to have a capacitance of $80\;pF$ is

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $(U )$ as $\varepsilon  = \alpha U$ where $\alpha  = 2{V^{ – 1}}$. A similar capacitor with no dielectric is charged to ${U_0} = 78\,V$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.