Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100\,V$ $D.C.$ source. Capacitor $C _1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C _2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $.........V.$ (Assuming Dielectric constant $=10$ )

A capacitor is connected to a $10\,V$ battery. The charge on the plates is $10\,\mu C$ when medium between plates is air. The charge on the plates become $100\,\mu C$ when space between plates is filled with oil. The dielectric constant of oil is

A capacitor has some dielectric between its plates and the capacitor is connected to a $\mathrm{D.C.}$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

How does the polarised dielectric modify the original external field inside it ?

A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is

Due to which the surface charge density arises on the surface of a dielectric slab, when it is placed in a uniform electric field ?