Adielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
the electric field in the capacitor
the charge on the capacitor
the potential difference between the plates
the stored energy in the capacitor.
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 :
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 ?