In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate \(=A\) )

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Answer & Solution

Correct Answer

(B) \(\frac{15}{34} \frac{{K\varepsilon}_{0} {A}}{{d}}\)

Detailed explanation

\(\frac{1}{C_{eq}}=\frac{d}{K \varepsilon_{0} A}+\frac{2 d}{3 K \varepsilon_{0} A}+\frac{3 d}{5 K \varepsilon_{0} A}\)

\(C_{eq}=\frac{15 K \varepsilon_{0} A}{34 d}\)

Same subject

List\(-I\)	List\(-II\)
\((a)\) \({R}_{{H}}\) (Rydberg constant)	\((i)\) \({kg} {m}^{-1} {s}^{-1}\)
\((b)\) \(h\) (Planck's constant)	\((ii)\) \({kg} {m}^{2} {s}^{-1}\)
\((c)\) \(\mu_{{B}}\) (Magnetic field energy density)	\((iii)\) \({m}^{-1}\)
\((d)\) \(\eta\) (coefficient of viscocity)	\((iv)\) \({kg} {m}^{-1} {s}^{-2}\)

From JEE Mains