A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure. 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 [figure]
- k = k1 + k2 + k3 + 3k4
- k = 23(k1 + k2 + k3) + 2k4
- 2k = 3k1 + k2 + k3 + 1k4
- 1k = 1k1 + 1k2 + 1k3 + 32k4
Answer
(3) 2k = 3k1 + k2 + k3 + 1k4
Solution
The upper half (thickness d2) has k1, k2, k3 side by side (each of area A3) these three are in parallel:
C123 = 0 (A/3)(k1 + k2 + k3)d/2
= 2 0 A(k1 + k2 + k3)3d
Lower half: C4 = 0 A k4d/2 = 2 0 A k4d
C123 and C4 are in series, and the single dielectric gives C = k 0 Ad:
dk 0 A = 3d2 0 A(k1 + k2 + k3) + d2 0 A k4
2k = 3k1 + k2 + k3 + 1k4
Answer: (3)