JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(=A\) )
- A \(\frac{25}{6} \frac{{K} \varepsilon_{0} {A}}{{d}}\)
- B \(\frac{15}{34} \frac{{K\varepsilon}_{0} {A}}{{d}}\)
- C \(\frac{15}{6} \frac{{K} \varepsilon_{0} {A}}{{d}}\)
- D \(\frac{9}{6} \frac{{K} \varepsilon_{0} {A}}{{d}}\)
Answer & Solution
Correct Answer
(B) \(\frac{15}{34} \frac{{K\varepsilon}_{0} {A}}{{d}}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{C_{eq}}=\frac{d}{K \varepsilon_{0} A}+\frac{2 d}{3 K \varepsilon_{0} A}+\frac{3 d}{5 K \varepsilon_{0} A}\) \(C_{eq}=\frac{15 K \varepsilon_{0} A}{34 d}\)
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