$A$ 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. (in $V$)

In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

The separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t$ $(t < d)$ is introduced between the plates,its capacitance becomes

$A$ parallel-plate capacitor is connected to a resistanceless circuit with a battery until the capacitor is fully charged. The battery is then disconnected from the circuit and the plates of the capacitor are moved to half of their original separation using insulated gloves. Let $V_{new}$ be the potential difference across the capacitor plates when the plates have moved. Let $V_{old}$ be the potential difference across the capacitor plates when they were connected to the battery. Find the ratio $\frac{V_{new}}{V_{old}}$.

The distance between the plates of a parallel plate capacitor is $d$. $A$ metal plate of thickness $d/2$ is introduced between the plates. The capacitance will then be

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?