The capacitance of a spherical capacitor in air is $50 \,\mu F$. When it is immersed in oil,its capacitance becomes $110 \,\mu F$. Calculate the dielectric constant of the oil.

The dielectric slab is introduced between the plates of a parallel plate charged capacitor. Which one of the following quantities will not change?

The distance between the plates of a charged parallel plate capacitor is $5 \ cm$ and the electric field inside the plates is $200 \ V/cm$. An uncharged metal bar of width $2 \ cm$ is fully immersed into the capacitor. The length of the metal bar is the same as that of the plate of the capacitor. The voltage across the capacitor after the immersion of the bar is......$V$

In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

An air capacitor has capacitance $C_1$. The space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The new capacitance of the capacitor is $C_2$. The ratio $\frac{C_1}{C_2}$ is ($d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the two dielectrics respectively).

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?