A capacitor has air as dielectric medium and two conducting plates of area 12 cm2 and they are 0.6 cm apart. When a slab of dielectric having area 12 cm2 and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given 0 = 8.834 10-12 F/m)
- 1.50
- 0.66
- 1.33
- 1
Answer
(1) 1.50
Solution
Initially C = 0 Ad, d = 0.6 cm
Since the slab raises C, the plate is moved away by 0.2 cm, so the gap now holds 0.6 cm of slab and 0.2 cm of air.
[figure]
C' = 0 A0.2 + 0.6K
C' = C 0.2 + 0.6K = 0.6
0.6K = 0.4 K = 1.5
Answer: (1)