When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on each plate. 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 magnitude 220 μC on each plate. What is the dielectric constant of the dielectric slab?

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valetta valetta · Answer 1 · 2019-07-31T08:49:46+02:00

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = [tex]\frac{(172\ \mu C)}{V}[/tex]

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = [tex]\frac{(172+220)\ \mu C)}{V}[/tex]

or

C' = [tex]\frac{(392)\ \mu C)}{V}[/tex]

now the dielectric constant (k) is given as:

[tex]k=\frac{C'}{C}[/tex]

substituting the values, we get

[tex]k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}[/tex]

or

k = 2.279