A parallel plate of capacitance $C_0$ is connected to a power supply $V_0$. A dielectric slab of dielectric constant $K=5$ is now inserted into the gap between the plates. (a) find the extra charge flown through the power supply and work done by the supply. (b) find change in energy stored in capacitor and heat produced in connecting wires.

A parallel plate capacitor of capacitance $100\mu F$ is connected a power supply of $200V$. A dielectric slab of dielectric constant 5 is now inserted into the gap between the plates. (a) Find the extra charge flown through the power supply and the work done by the supply. (b) Find the change in the electrostatic energy of the electric field in the capacitor.

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant $K=\frac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be :

An air filled parallel plate capacitor of capacitance $50\mu F$ is connected to a battery of 100 V. A slab of dielectric constant 4 is inserted in it to fill the space completely . Find the extra charge flown through the battery till it attains the steady state.

A capacitor of capacitance $200\mu F$ is connected to a 200V battery. Dielectric material of dielectric constant 2 is introduce in between the plates of the capacitor. Find change in energy stored in the process.

Two identical parallel plate capacitors are connected in series to a battery of $100V$. A dielectric slab of dielectric constant $4.0$ is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

A parallel capacitor of capacitance $C$ is charged and disconnected from the battery. The energy stored in it is $E$. If a dielectric slab of dielectric constant $6$ is inserted between the plates of the capacitor then energy and capacitance will become.

A parallel plate capacitor of capacitance $200\mu F$ is connected to a battery of 200 V. A dielectric slab of dielectric 2 is now inserted into the space between plates of capacitor while the battery remain connected .The change in the electrostatic energy in the capacitor will be __________J.

A parallel -plate air capacitor has capacity $C$. A dielectric slab of dielectric constant $K$, whose thickness is half of the air gap between the plates, is now inserted between the plates. The capacity of the capacitor will now be

The plates of a parallel plate capacitor with no dielectirc are connected to a volatage source. Now a dielectric of dielectric constant $K$ is inserted to fill the whose space between the plates with voltage source remainign connected to the capacitor-

The plates of a capacitor are connected to a source of emf and the energy stored is $U_0$. When dielectric material dielectric constant $K$ is introduced between the plates filling the whole space, the energy stored becomes $U$ Now.