The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness 38d and d2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1 = 1.25K2, the value of K1 is:
- 1.60
- 1.33
- 2.66
- 2.33
Answer
(3) 2.66
Solution
With slabs: C = 0 Ad - t1 - t2 + t1K1 + t2K2, and C = 2C0 = 2 0 Ad
d8 + 3d8K1 + d2K2 = d2
38K1 + 12K2 = 38
With K2 = K11.25: 12K2 = 1.252K1 = 58K1
38K1 + 58K1 = 1K1 = 38
K1 = 83 2.66
Answer: (3)