A parallel plate capacitor has charge 5 10-6 C. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4 10-6 C then the dielectric constant of the slab is .
Answer
5
Solution
Induced charge on the face of a dielectric slab: qi = q (1 - 1K )
4 10-6 = 5 10-6 (1 - 1K )
1 - 1K = 45
1K = 15
K = 5
Answer: 5