When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates,then the energy stored:

$A$ parallel plate capacitor is made of two plates of length $l$,width $w$ and separated by distance $d$. $A$ dielectric slab (dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{\partial U}{\partial x}$ where $U$ is the energy of the capacitor when the dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$,then the force on the dielectric when it is near the edge is

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:

$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

$A$ slab of material with a dielectric constant of $3$ has the same area as the plates of a parallel plate capacitor but has a thickness of $\left(\frac{3}{4}\right) d$,where $d$ is the separation between the plates. What is the electrical potential difference between the plates when the slab is inserted? The initial electrical potential difference is $V_0$.

When a dielectric material is introduced between the plates of a charged capacitor,the electric field between the plates: