A dielectric slab of relative premittivity (i.e. dielectric constant) 6 is introduced between the two plates of an 8`mu`F air capacitor, in order to completely occupy the space between the two plates. Find the new capacitance of the capacitor.

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab having dielectric constant (relative permittivity) K, is now introduced between its two plates in order to occupy the space completely. State, in terms of K, its effect on the following: The capacitance of the capacitor

An isolated 16 muF parallel plate air capacitor has a potential difference of 100 V. dielectric slab having relative permittivity (i.e., dielectric constant) = 5 is introduced to fill the space between the two plates completely. Calculate. (i) the new capacitance of the capacitor (ii) the new potential differece between the two plates of capacitor.

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is now introduced between the two plates to occupy the space completely. State the effect on the following: (i) the capacitance of the capacitor. (ii) potential difference between the plates. (iii) the energy stored in the capacitor.

In the arrangement shown, the battery is disconnected and a dielectric slab of dielectric constant K is insereted between the plates of capacitor with capacitance C, so as to completely fill the space. What is the final potential differnece across capacitor with capacitance 2C ?

A parallel - plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected . A slab of dielectric constant K is then inserted between the plate of the capacitor so as to fill the space between the plate .Find the work done on the system in the process of inserting the slab.

An air-cored parallel plate capacitor having a capacitance C is charged by connecting it to a battery. Now, it is disconnected from the battery and a dielectric slab of dielectric constant K is inserted between its plates so as to completely fill the space between the plates. Compare : (I) Initial and final capacitance (ii) Initial and final charge. (iii) Initial and final potential difference. (iv) Initial an final electric field between the plates. (v) Initial and final energy stored in the capacitor.

A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected ? Justify your answer.

A parallel plate capacitor (without dielectric) is charged by a battery and kept connected to the battery. A dielectric salb of dielectric constant 'k' is inserted between the plates fully occupying the space between the plates. The energy density of electric field between the plates will:

When a dielectric slab is inserted between the plates of one of the two identical capacitors shown in the figure then match the following: