$A$ capacitor $C$ is charged to a potential difference $V$ and the battery is disconnected. Now,if the capacitor plates are brought closer slowly by some distance,then:

$A$ parallel plate capacitor is constructed with two plates. Each plate has an area of $100 \ cm^2$ and they are separated by a distance of $1 \ mm$. $A$ dielectric with a dielectric constant of $5.0$ and a dielectric strength of $1.9 \times 10^7 \ V/m$ is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing dielectric breakdown.

Between the plates of a parallel plate capacitor of plate area $A$ and capacity $0.025 \mu F$,a metal plate of area $A$ and thickness equal to $1/3$ of the separation between the plates of the capacitor is introduced. If the capacitor is charged to $100 \ V$,then the amount of work done to remove the metal plate from the capacitor is (in $\mu J$)

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

Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.
Reason : The surface density of charge on the plate remains constant or unchanged.

$A$ simple pendulum of mass $m$, length $l$, and charge $+q$ is suspended in the electric field produced by two conducting parallel plates as shown. The value of the deflection of the pendulum in the equilibrium position will be -