Question

When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of 200 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 250 µC on each plate. What is the dielectric constant of the dielectric slab?

Answer 1

Answer:

The value is $\epsilon = 2.25$

Explanation:

From the question we are told that

The charge acquired on each plate is $q = 200 \mu C = 200 *10^{-6} \ C$

The additional charge accumulated on each plate is $q_a = 250 \mu C = 250 *10^{-6} \ C$

Generally the capacitance of the capacitor before a dielectric slab is inserted is

$C_i = \frac{q}{V}$

=> $C_i = \frac{200 \mu C}{V}$

Generally the total charge on each plate of the capacitor when the dielectric slab was inserted is

$Q = q + q_a$

=> $Q = ( 200 + 250 ) \mu C$

=> $Q = 450 \mu C$

Generally the capacitance of the capacitor after a dielectric slab is inserted is mathematically represented as

$C = \frac{450 \mu C}{V}$

Generally the dielectric constant is mathematically represented as

$\epsilon = \frac{C}{C_i}$

=>      $\epsilon =  \frac{450 \mu C}{200  \mu C}$

=>       $\epsilon =  2.25$

Answer 2

The dielectric constant of the dielectric slab is 2.25.

Calculation of the  dielectric constant:

Since the charge on each playe should be q = 200 µC = 200*10^-6C

Now the additional charge accumulated should be

qa = 250µC = 250*10^-6C

Now we know that

Ci = q/v

Ci = 200µC/ V

Also,

Q = q+ qa

= 9200 + 250)µC

= 450µC

Now C = 450µC /  V

So here the dielectric constant should be

= 450µC/200µC

= 2.25

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