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

  • A

    the electric field in the capacitor

  • B

    the charge on the capacitor

  • C

    the potential difference between the plates

  • D

    the stored energy in the capacitor.

Similar Questions

A parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$ its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is

A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E _{1}$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _{2}$. The ratio $E _{1} / E _{2}$ will be :

  • [JEE MAIN 2022]

A capacitor of capacity $'C'$ is connected to a cell of $'V'\, volt$. Now a dielectric slab of dielectric constant ${ \in _r}$ is inserted in it keeping cell connected then

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.......$\mu F$

What is polarisation ?