67.A parallel plate capacitor having plate area 400cm2 and separation between the plates 1mm is connected to a power supply of 100V.A dielectric slab of thickness 0.5 mm and dielectric constant 5 is inserted into the gap . if the power supply is disconnected and the dielectric slab is taken out , find the increase in energy

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Q. A parallel-plate capacitor having plate area 400 cm² and separation between the plates 1⋅0 mm is connected to a power supply of 100 V. A dielectric slab of thickness 0⋅5 mm and dielectric constant 5⋅0 is inserted into the gap. (a) Find the increase in electrostatic energy. (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out?

Q. A parallel plate capacitor of capacitance

100 μ F

is connected to a power supply of

200 V

. A dielectric slab of dielectric constant

5

is now inserted into the gap between the plates. The workdone by power supply will be:

The plates of a parallel plate capacitor are charged to $200 V$ and then, the charging battery is disconnected. Now, a dielectric slab of dielectric constant $5$ and thickness $4 m m$ is inserted between the capacitor plates. To maintain the original capacity, the increase in the separation between the plates of the capacitor is:

Q. Space between the plates of a parallel plate capacitor is filled with a dielectric slab. The capacitor is charged and then the supply is disconnected to it. If the slab is now taken out , then:

A parallel-plate capacitor of capacitance $5 μ F$ is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination (d) How much charge has flown through the battery after the slab is inserted ?

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