Q. 594.8( 6 Votes )

# 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 plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.

Answer :

Before inserting slab-

We know, capacitance c is given by-

Where,

A= Plate Area

d= separation between the plates,

∈_{0} = Permittivity of free space = 8.854 × 10^{-12} m^{-3} kg^{-1} s^{4} A^{2}

Energy stored by the capacitor

where C is the capacitance and V is the applied voltage.

After inserting slab capacitance c is given by-

Where,

A= Plate Area

d= separation between the plates,

∈_{0} = Permittivity of free space = 8.854 × 10^{-12} m^{-3} kg^{-1} s^{4} A^{2}

k = dielectric strengthof the material

Also, the final voltage becomes

Energy stored by the capacitor–

where ,

C is the capacitance and V is the applied voltage, k is the dielectric constant of the material

The work done on the system in the process of inserting the slab

= change in energy

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A parallel-plate capacitor has plate area 20 cm^{2}, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy for the capacitor 8.9 μs after the connections are made.

How many time constants will elapse before the energy stored in the capacitor reaches half of its equilibrium value in a charging RC circuit?

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