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Question: Answered & Verified by Expert
A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?

PhysicsCapacitanceJEE Main
Options:
  • A $\frac{3 K \varepsilon_{0} A}{4 d}$
  • B $\frac{4 K \varepsilon_{0} A}{3 d}$
  • C $\frac{(K+1) \varepsilon_{0} A}{2 d}$
  • D $\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$
Solution:
2514 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 $C_{3}$ )

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