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

  $\frac{3 K \varepsilon_{0} A}{4 d}$
• B

  $\frac{4 K \varepsilon_{0} A}{3 d}$
• C

  $\frac{(K+1) \varepsilon_{0} A}{2 d}$
• Correct answer
  D

  $\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$

Solution

$$ c_{1}=\frac{(A / 2) \varepsilon_{0}}{d / 2}=\frac{A \varepsilon_{0}}{d}, c_{2}=K \frac{A \varepsilon_{0}}{d}, c_{3}=K \frac{A \varepsilon_{0}}{2 d} $$
$$ \therefore \quad c_{e q .}=\frac{c_{1} \times c_{2}}{c_{1}+c_{2}}+c_{3}=\frac{(3+\mathrm{K}) \mathrm{K} A \varepsilon_{0}}{2 d(\mathrm{~K}+1)} $$ $\left(\because C_{1}\right.$ and $C_{2}$ are in series and resultant of these two in parallel with $C_{3}$ )

Hence, the correct answer is option: 'D'.