The capacity of a parallel plate capacitor is $5 \mu F$. When a glass plate is placed between the plates of the capacitor,its potential difference reduces to $1/8$ of the original value. The magnitude of the relative dielectric constant of the glass is

$A$ parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$. What will be the capacitance if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

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 has an area of $6 \, cm^2$ and a separation of $3 \, mm$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$,and $K_3 = 14$. The dielectric constant of a material which,when fully inserted in the above capacitor,gives the same capacitance would be:

Two identical capacitors $1$ and $2$ are connected in series. Capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0 \text{ volts}$. The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charges stored in the capacitors before removing the slab,and $Q'_1$ and $Q'_2$ be the values after removing the slab. Then:

If a capacitor with air,a capacitor with a dielectric,and a capacitor with a conducting slab have capacitances $C_1$,$C_2$,and $C_3$ respectively,then: