Question

22.
The space between the plates of a parallel plate capacitor is completely
filled in two ways. In the first case, it is filled with a slab of dielectric
constant K. In the second case, it is filled with two slabs of equal
thickness and dielectric constants K1 and K2 respectively as shown in the
figure. The capacitance of the capacitor is same in the two cases. Obtain
the relationship between K, K1 and K2.
K1
K2
d-
No
(Case 1)
(Case 2)​

Answer 1

Answer:

K=((K1×K2)/K1×K2)

Explanation:

Answer 2

 $K=2\frac{K_1K_2}{K_1+K_2}$    

Explanation:

The capacitance of the capacitor filled with a dielectric slab (K) is given by:

$C=\frac{K\epsilon_oA}{d}$

where, A is the area of the plate, K is the dielectric constant, d is the thickness of the plate.

When the capacitor is filled with two slabs of equal thickness, the capacitance is as follows:

$C=\frac{2\epsilon_o A}{d} (\frac{K_1K_2}{K_1+K_2})$

Area of the plates remain same and thickness of each slab is d/2.

It is given that the capacitance is equal. So, equate the above two formulas:

$\frac{K\epsilon_oA}{d}=\frac{2\epsilon_o A}{d} (\frac{K_1K_2}{K_1+K_2})\\K=2\frac{K_1K_2}{K_1+K_2}$

Learn more about dielectrics and capacitors:

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