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JEE Advanced 2019
Paper-1 2019
Question
A parallel plate capacitor of capacitance C has spacing d between two plates having area A . The region between the plates is filled with N dielectric layers, parallel to its plates, each with thickness $\delta=\frac{\mathrm{d}}{\mathrm{N}}$. The dielectric constant of the $\mathrm{m}^{\text {th }}$ layer is $\mathrm{K}_{\mathrm{m}}=\mathrm{K}\left(1+\frac{\mathrm{m}}{\mathrm{N}}\right)$. For a very large $\mathrm{N}\left(>10^3\right)$, the capacitance C is $\alpha\left(\frac{\mathrm{K} \in_0 \mathrm{~A}}{\mathrm{~d} / \mathrm{n} 2}\right)$. The value of $\alpha$ will be $\_\_\_\_$ .
$\left[\epsilon_0\right.$ is the permittivity of free space $]$
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Question Tags
JEE Advance
Physics
Easy
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