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Question
A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d1 and dielectric constant k1 and the other has thickness d2 and dielectric constant k2 as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (= d1 + d2) and effective dielectric constant k. The k is ______.
Options
`(k_1d_1 + k_2d_2)/(d_1 + d_2)`
`(k_1d_1 + k_2d_2)/(k_1 + k_2)`
`(k_1k_2 (d_1 + d_2))/((k_1d_1 + k_2d_2))`
`(2k_1k_2)/(k_1 + k_2)`
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Solution
A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d1 and dielectric constant k1 and the other has thickness d2 and dielectric constant k2 as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (= d1 + d2) and effective dielectric constant k. The k is `underline((k_1k_2 (d_1 + d_2))/((k_1d_1 + k_2d_2)))`.
Explanation:
Here the system can be considered as two capacitors C1 and C2 connected in series as shown in figure.
The capacitance of parallel plate capacitor filled with dielectric block has thickness d1 and dielectric constant K2 is given by
`1/C_(eq) = 1/C_1 + 1/C_2` ⇒ `C_(eq) = (C_1C_2)/(C_1 + C_2)`
`C_(eq) = ((k_1ε_0A)/(d_1))/((k_1ε_0A)/(d_1)) ((k_2ε_0A)/(d_2))/((k_2ε_0A)/(d_2)) = (k_1k_2ε_0A)/(k_1d_2 + k_2d_1)` ......(i)
We can write the equivalent capacitance as `C = (kε_0A)/(d_1 + d_2)` ......(ii)
On comparing (i) and (ii) we have
`k = (k_1k_2(d_1 + d_2))/(k_1d_2 + k_2d_1)`
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