Adielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

A parallel plate capacitor is made of two circular plates separated by a distance $5\ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4$ $ Vm^{-1}$ the charge density of the positive plate will be close to

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant $K $ is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

A parallel plate capacitor with air as medium between the plates has a capacitance of $10\,\mu F$. The area of capacitor is divided into two equal halves and filled with two media as shown in the figure having dielectric constant ${k_1} = 2$and ${k_2} = 4$. The capacitance of the system will now be.......$\mu F$

In a parallel plate capacitor set up, the plate area of capacitor is $2 \,m ^{2}$ and the plates are separated by $1\, m$. If the space between the plates are filled with a dielectric material of thickness $0.5\, m$ and area $2\, m ^{2}$ (see $fig.$) the capacitance of the set-up will be $.........\, \varepsilon_{0}$

(Dielectric constant of the material $=3.2$ ) and (Round off to the Nearest Integer)