Question

Two thin dielectric slabs of dielectric constants K₁ and K₂ (K₁ < K₂) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field 'E' between the plates with distance 'd' as measured from plate P is correctly shown by:

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is B

To solve this problem, we need to understand how the electric field varies across a parallel plate capacitor with dielectrics of different dielectric constants inserted.

The electric field E in a region with a dielectric constant K is given by:

E = \frac{E_0}{K}

where E_0 is the electric field in the absence of the dielectric.

Given:

• Dielectric constant of slab 1: K_1
• Dielectric constant of slab 2: K_2
• K_1 < K_2

Let's analyze the electric field variation:

1. In the region with no dielectric, the field is E_0.
2. In the region with dielectric slab K_1, the electric field is reduced to E_1 = \frac{E_0}{K_1}. Since K_1 < K_2, E_1 will be greater than E_2 in the slab with K_2.
3. In the region with dielectric slab K_2, the electric field is E_2 = \frac{E_0}{K_2}.

Thus, the field is strongest in the region with lower dielectric constant, K_1, and weakest in the region with higher dielectric constant, K_2.

Therefore, the correct graph would show a decrease in the electric field from the region with no dielectric to the region with dielectric K_1, and a further decrease to the region with dielectric K_2. The correct option is represented by: