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JEE Main 2022 (28 Jul, Shift 1) · Dielectrics

Dielectrics question from JEE Main 2022 (28 Jul, Shift 1)

Q68JEE Main 2022 · 28 JUL · SHIFT 1Medium

Dielectrics

Two capacitors, each having capacitance $40\ \mu\text{F}$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\ \mu\text{F}$ . The value of $K$ will be:

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Answer: (a)

Series: $\frac{40K \cdot 40}{40K + 40} = 24 \Rightarrow \frac{40K}{K+1} = 24$ .
$40K = 24K + 24 \Rightarrow 16K = 24 \Rightarrow K = 1.5$
So the answer is (a) 1.5.

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