Introduction of a dielectric slab of dielectric constant K between the plates |When the battery is disconnected |When the battery remains connected |Questions |Force between capacitor plates

Two identical capacitors A and B shown in the given circuit are joined in series with a battery. If a dielectric slab of dielectric constant K is slipped between the plates of capacitor B and battery remains connected, then the energy of capacitor A will-

A parallel plate capacitor having capacitance C farad is connected with a battery of emf V volts. Keeping the capacitor cannected with the battery, a dielectric slab of dielectric constant K is inserted between the plates. The dimensions of the slab are such that it fills the space between the capacitor plates. Then,

When a dielectric material is introduced between the plates of a charged condenser, after disconnected the battery the electric field between the plates

Find an expression for the capacitance of a parallel plate capacitor when a dielectric slab of dielectric constant K and thickness t=(d)/(2) but the same area on the plates is inserted between the the capacitors plate. (d=separation between the plates)

A parallel plate capacitor of capacitance C_(0) is charged with a charge Q_(0) to a potential difference V_(0) and the battery is then disconnected Now a dielectric slab of dielectric constant k is inserted between the plates ofcapacitor. The dimensions of the slab are such that it completely fills the space between the plates, then :

A parallel plate capacitor is connecyed to a battery of emf V volts as shown. 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.

Figure shows a parallel plate capacitor with plate area A and plate separation d . A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dieletric constant K is placed in between the plates of the capacitor as shown. The electric field in the gaps between the plates and the electric slab will be

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 parallel plate capacitor of capacitance 200muF 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.

Two identical capacitors are connected as shown. A dielectric slab Is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. Compare the capacitance, the charge, potential difference and stored energy of each capacitor before and arter introducting of dielectric slab.