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A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?
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1917 Upvotes
Verified Answer
The correct answer is:
$\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$
$$
c_{1}=\frac{(A / 2) \varepsilon_{0}}{d / 2}=\frac{A \varepsilon_{0}}{d}, c_{2}=K \frac{A \varepsilon_{0}}{d}, c_{3}=K \frac{A \varepsilon_{0}}{2 d}
$$
$$
\therefore \quad c_{e q .}=\frac{c_{1} \times c_{2}}{c_{1}+c_{2}}+c_{3}=\frac{(3+\mathrm{K}) \mathrm{K} A \varepsilon_{0}}{2 d(\mathrm{~K}+1)}
$$
$\left(\because C_{1}\right.$ and $C_{2}$ are in series and resultant of these two in parallel with $C_{3}$ )
c_{1}=\frac{(A / 2) \varepsilon_{0}}{d / 2}=\frac{A \varepsilon_{0}}{d}, c_{2}=K \frac{A \varepsilon_{0}}{d}, c_{3}=K \frac{A \varepsilon_{0}}{2 d}
$$
$$
\therefore \quad c_{e q .}=\frac{c_{1} \times c_{2}}{c_{1}+c_{2}}+c_{3}=\frac{(3+\mathrm{K}) \mathrm{K} A \varepsilon_{0}}{2 d(\mathrm{~K}+1)}
$$
$\left(\because C_{1}\right.$ and $C_{2}$ are in series and resultant of these two in parallel with $C_{3}$ )
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