A parallel plate capacitor of area A , plate separation d and capacitance C is filled with three dif

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1. Electric Charges and Fields

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A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constant $K_1,K_2$ and $K_3$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K$ is given by: ($A =$ Area of plates)

$\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}} + \frac{1}{{2{K_3}}}$

$\frac{1}{K} = \frac{1}{{{K_1} + {K_2}}} + \frac{1}{{2{K_3}}}$

$K = \frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}} + 2{K_3}$

$K = K_1 + K_2 + 2K_3$

Solution

Let

$\mathrm{C}_{1} \frac{\varepsilon_{0}(\mathrm{A} / 2) \mathrm{K}_{1}}{(\mathrm{d} / 2)}, \mathrm{C}_{2}=\frac{\varepsilon_{0}(\mathrm{A} / 2) \mathrm{K}_{2}}{(\mathrm{d} / 2)}, \mathrm{C}_{3}=\frac{\varepsilon_{0} \mathrm{A}\left(\mathrm{K}_{3}\right)}{(\mathrm{d} / 2)}$

Then, $\mathrm{C}=\frac{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{C}_{3}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{A}}{\mathrm{d}}$

i.e., $\quad \frac{1}{\mathrm{K}}=\frac{1}{\mathrm{K}_{1}+\mathrm{K}_{2}}+\frac{1}{2 \mathrm{K}_{3}}$

Standard 12

Physics

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Solution

Similar Questions

A charge $2\,\mu C$ is taken from infinity to a point in an electric field, without changing its velocity. If work done against forces is $20\,\mu J$ then potential at that point will be…..$V$

The four capacitors, each of $25\,\mu F$ are connected as shown in Fig. The $dc$ voltmeter reads $200\,V$. The charge on each plate of capacitor is

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