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Question

A parallel plate capacitor is filled equally(half) with two dielectrics of dielectric constants $\varepsilon_1$ 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 :

First Configuration 

 

The correct answer is

$\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}$

To solve the problem, we need to find the ratio of capacitances in two configurations: the first configuration with the dielectrics stacked vertically and the second configuration with the dielectrics placed side by side.

Configuration Analysis

First Configuration

In the first configuration, the capacitor is filled with dielectrics of constants \(\varepsilon_1\) and \(\varepsilon_2\) separated vertically, forming two capacitors in series.

The capacitance for a series arrangement is given by:

\[\frac{1}{C_1} = \frac{1}{C_{1_1}} + \frac{1}{C_{1_2}}\]

Where:

  • \(C_{1_1} = \frac{\varepsilon_1 \varepsilon_0 A}{2d}\) for the section with dielectric \(\varepsilon_1\)
  • \(C_{1_2} = \frac{\varepsilon_2 \varepsilon_0 A}{2d}\) for the section with dielectric \(\varepsilon_2\)

Combining these, we get:

\[\frac{1}{C_1} = \frac{2d}{\varepsilon_1 \varepsilon_0 A} + \frac{2d}{\varepsilon_2 \varepsilon_0 A}\]\[C_1 = \frac{\varepsilon_1 \varepsilon_2 \varepsilon_0 A}{d(\varepsilon_1 + \varepsilon_2)}\]

Second Configuration

In the second configuration, the capacitor consists of two capacitors in parallel.

The capacitance for a parallel arrangement is given by:

\[C_2 = C_{2_1} + C_{2_2}\]

Where:

  • \(C_{2_1} = \frac{\varepsilon_1 \varepsilon_0 \left(\frac{A}{2}\right)}{d}\)
  • \(C_{2_2} = \frac{\varepsilon_2 \varepsilon_0 \left(\frac{A}{2}\right)}{d}\)
\[C_2 = \frac{\varepsilon_0 A}{2d} (\varepsilon_1 + \varepsilon_2)\]

Ratio of Capacitances

The ratio of the capacitances is:

\[\frac{C_1}{C_2} = \frac{\frac{\varepsilon_1 \varepsilon_2 \varepsilon_0 A}{d(\varepsilon_1 + \varepsilon_2)}}{\frac{\varepsilon_0 A}{2d}(\varepsilon_1 + \varepsilon_2)}\]\[\frac{C_1}{C_2} = \frac{2\varepsilon_1 \varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2} \] \]\]

The given correct answer matches this result:

\(\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}\)

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Important Questions from Electricity and Magnetism

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