Question

A parallel plate capacitor of area '$A$' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?

• A. $\frac{3K\varepsilon _0\:A}{4d}\:$
• B. $\frac{3K\varepsilon _0\:A}{2d}\:$
• C. $\frac{(K+1) \varepsilon_{0} A}{2 d}$
• D. $\frac{K(K + 3)\varepsilon_0 A}{2(K + 1 )d}$

Answer 1

Answer: (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 \epsilon_{0}}{2 d} $
$\therefore c_{\text {eq. }}=\frac{c_{1} \times c_{2}}{c_{1}+c_{2}}+c_{3}=\frac{(3+K) K A \varepsilon_{0}}{2 d(K+1)}$
$\left(\because C_{1}\right.$ and $C_{2}$ are in series and resultant of these two in parallel with $\left.C_{3}\right)$.