Question

When a certain air-filled parallel-plate capacitor is connected cross a battery, it acquires a charge (on each plate) of 160 µC. While the battery connection is maintained, a dielectric slab is inserted into and fills the region between the plates. This results in the accumulation of an additional charge of 220 µC on each plate. What is the dielectric constant of the dielectric slab?

Answer 1

Answer:

k = 2.45

Explanation:

It is given that,

Charge acquired on each plate of the capacitor due to a battery, $q=160\ \mu C$

If V is the potential difference across the plates, capacitance in air is given by :

$C_a=\dfrac{q}{V}=\dfrac{160}{V}$.............(1)

Let k is the dielectric constant of the dielectric slab. While the battery connection is maintained, a dielectric slab is inserted into and fills the region between the plates. This results in the accumulation of an additional charge of 220 µC on each plate.

New charge becomes, $Q=220\ \mu C+220\ \mu C=440\ \mu C$

$C_d=\dfrac{Q}{V}=\dfrac{440}{V}$.............(2)

The dielectric constant of the dielectric slab is given by :

$k=\dfrac{C_d}{C_a}$

$k=\dfrac{440}{160}$

k = 2.45

So, the dielectric constant of the dielectric slab is 2.45. Hence, this is the required solution.