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 is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected ? Justify your answer in each case.

An uncharged parallel plate capacitor is connected to a battery. The electric field between the plates is 10V/m. Now a dielectric of dielectric constant 2 is inserted between the plates filling the entries space. The electric field between the plates now is

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:

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

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 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.

Answer the following: (a) When a parallel plate capacitor is connected across a dc battery, explain briefly how the capacitor gets charged. (b) A parallel plate capacitor of capacitance 'C' is charged to 'V' volt by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant 1ltklt2 is introduced to fill the space between the plates. How will the following be affected ? (i) The electric field between the plates of the capacitor. (ii) The energy stored in the capacitor. Justify your answer in each case. (c) The electric potential as a function of distance 'x' is shown in the figure. Draw a graph of the electric field E as a function of x.

A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,

An air capacitor is first charged through a battery. The charging battery is then removed and a electric slab of dielectric constant K=4 is inserted between the plates. Simultaneously, the distance between the plates is reduced to half, then find change, C, E, V and U .

A parallel plate capacitor is connected to a battery of emf V volts. Now a slab of dielectric constant k=2 is inserted between the plates of capacitor without disconnecting the battery. The electric field between the plates of capacitor after inserting the slab is E= (PV)/(2d) . Find the value of P.