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

Concept:

Let the two plates are kept parallel to each other separated by a distance 'd' and the cross-sectional area of each plate is A.

Electric field by a single thin plate (assuming it to be infinite) is:

\(E' = \frac{\sigma }{{2{\varepsilon _o}}}\)

The electric field because of both the plates will be the sum of the two, i.e.

\(E = \frac{\sigma }{{{\varepsilon _o}}}\)

Since \({\sigma = \frac{Q}{A}} \)

\(E = \frac{Q}{{A{\varepsilon _0}}}\)

Important Observations:

• The electric field is inversely proportional to the dielectric constant. 

           \({E_{medium}} = \frac{{{E_{vacuum}}}}{K}\)

• The electric field inside the dielectrics will be less than the electric field in vacuum.
• The electric field inside the dielectric could not be zero.
• As K₂ > K₁ the drop in electric field for K₂ dielectric must be more than K₁.

Conclusion: From the above observations, we can conclude that the electric field will be maximum and constant in portions without a dielectric. Inside the dielectric the drop in electric field will be more for thick dielectric, i.e.for K₂. The plot of Option (3) is, therefore, correct.