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 [figure]
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Answer
(3) [figure]
Solution
In air between the plates: E0 = 0
Inside a dielectric: E = E0K, so the field drops inside each slab (it is not zero).
K1 < K2 E0K1 > E0K2, i.e. the drop is smaller in K1 and larger in K2
Between the slabs (air) the field returns to E0.
Graph (3) is correct.
Answer: (3)