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# A capacitor stores 50 μC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100 μC flows through the battery. Find the dielectric constant of the material inserted.

Answer :

Given-

Initially, the charge on the capacitor = 50 μC

Now, let the dielectric constant of the material inserted in the gap be *k*.

When this dielectric material is inserted, 100 μC of extra charge flows through the battery

Therefore , the net charge on the capacitor becomes

50 + 100 = 150 μC

Now, we know capacitance of a material is given by –

Where q is charge on the capacitance and v is the applied voltage

Also

Where A is the plate area and ∈_{0} is the permittivity of the free space.

Initially, without dielectric material inserted, capacitance is given by

1)

Similarly, with the dielectric material place, capacitance is given by

2)

On dividing *1*) by *2*), we get

⇒

Thus, the dielectric constant of the given material is 3.

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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.

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HC Verma - Concepts of Physics Part 2