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 _______________.

Solution

Physics sir · 4 steps

The slab's induced charge cancels part of the plate charge, and what is left over sets K.

1. What is left. The induced charge on the slab face has the opposite sign to the charge on the plate next to it. Inside the slab, the field comes only from the net charge Q - Qᵢ, not from the full Q.
2. Link to K. A dielectric cuts the field to 1/K of its value without the slab. So the net charge must be Q/K, which gives the standard result below.

   Qᵢ = Q(1 - 1/K)

3. Put in numbers. 4 × 10⁻⁶ = 5 × 10⁻⁶(1 - 1/K), so 1 - 1/K = 4/5 and 1/K = 1/5.
4. So. The dielectric constant of the slab is K = 5.

The common slip: 1.25. Taking K as the plain ratio Q/Qᵢ = 5/4 gives 1.25. The ratio that equals K is Q/(Q - Qᵢ) = 5/1.

The idea to keep. Induced charge on a dielectric is Q(1 - 1/K). The net charge Q - Qᵢ = Q/K is what actually makes the field inside the slab.

Answer from the NTA final answer key. Solution by PrepWiser.