$A$ parallel plate capacitor has a plate separation of $0.01\, mm$ and uses a dielectric (whose dielectric strength is $19\, kV/mm$) as an insulator. The maximum potential difference that can be applied to the terminals of the capacitor is......$V$

$A$ parallel plate capacitor with plate separation $5 \text{ mm}$ is charged by a battery. On introducing a mica sheet of $2 \text{ mm}$ thickness and maintaining the connections of the plates with the terminals of the battery, it is found that it draws $25 \%$ more charge from the battery. The dielectric constant of mica is . . . . . . .

$A$ 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 -

Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is:

Two dielectric slabs of dielectric constants $K_1$ and $K_2$ and of the same thickness are inserted in a parallel plate capacitor. Given $K_1 = 2K_2$. If the potential differences across the slabs are $V_1$ and $V_2$ respectively,then:

$A$ parallel plate capacitor having capacitance $C$ farad is connected to a battery of emf $V$ volt. Keeping the capacitor connected to the battery,a dielectric slab of dielectric constant $K$ is inserted between the plates. The dimensions of the slab are such that it fills the whole space between the capacitor plates. Then $:-$