JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant \(\varepsilon_1\) and \(\varepsilon_2\), as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are \(C_1\) and \(C_2\) respectively, then \(\frac{C_1}{C_2}\) is :


- A \(\frac{\varepsilon_1 \varepsilon_2^2}{\left(\varepsilon_1+\varepsilon_2\right)^2}\)
- B \(\frac{4 \varepsilon_1 \varepsilon_2}{\left(\varepsilon_1+\varepsilon_2\right)^2}\)
- C \(\frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1+\varepsilon_2}\)
- D \(\frac{\varepsilon_0\left(\varepsilon_1+\varepsilon_2\right)}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{4 \varepsilon_1 \varepsilon_2}{\left(\varepsilon_1+\varepsilon_2\right)^2}\)
Step-by-step Solution
Detailed explanation
Area of plate is \(A\). then…
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