A parallel plate capacitor of capacitance `100muF` is connected a power supply of `200 V`. 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 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.

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 parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K= (5)/(3) is inserted between the plates, the magnitude of the induced charge will be :

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.

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.

A parallel-plate air capacitor of plate separation 2mm and capacitance 1muF is charged to 100V. A dielectric slab of relative permittivity 50 is now inserted so as to fill the space between the plates. (i)Find the polarisation charge on one of the boundaries of the dielectric slab. (ii) Find the magnitude of the polarisation of the dielectric slab. (epsilon_(0) = 8.85 xx 10^(-12) C^(2)//N.m^(2))

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.

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