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Two parallel plates of area A are separated by two different dielectrics as shown in figure. The net capacitance is
Two parallel plates of area A are separated by two different dielectrics as shown in figure. The net capacitance is
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Two parallel plates of area A are separated by two different dielectrics as shown in figure. The net capacitance is \includegraphics[width=0.5\linewidth]49phy.png \labelfig:placeholder
Updated On: Jun 20, 2026
- $\frac{\epsilon_{0}A}{2d}$
- $\frac{\epsilon_{0}A}{d}$
- $\frac{3\epsilon_{0}A}{d}$
- $\frac{4\epsilon_{0}A}{3d}$
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The Correct Option is D
Solution and Explanation
Step 1: Formula
Net capacitance for dielectrics in series: $C = \frac{\epsilon_{0}A}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}}$.
Step 2: Values
$d_{1} = d/2, K_{1} = 1$ and $d_{2} = d/2, K_{2} = 2$.
Step 3: Calculation
$C = \frac{\epsilon_{0}A}{\frac{d}{2}(\frac{1}{1} + \frac{1}{2})} = \frac{\epsilon_{0}A}{\frac{d}{2}(\frac{3}{2})} = \frac{4\epsilon_{0}A}{3d}$.
Hence, the Answer is: (d)
Net capacitance for dielectrics in series: $C = \frac{\epsilon_{0}A}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}}$.
Step 2: Values
$d_{1} = d/2, K_{1} = 1$ and $d_{2} = d/2, K_{2} = 2$.
Step 3: Calculation
$C = \frac{\epsilon_{0}A}{\frac{d}{2}(\frac{1}{1} + \frac{1}{2})} = \frac{\epsilon_{0}A}{\frac{d}{2}(\frac{3}{2})} = \frac{4\epsilon_{0}A}{3d}$.
Hence, the Answer is: (d)
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