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Two thin dielectric slabs of dielectric constants `K_(1)` and `K_(2) (K_(1) lt K_(2))` 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

Text Solution

Verified by Experts

The correct Answer is:
C

Electric field inside parallel plate capacitor having charge `Q` at place where dielectric is absent `= (Q)/(A epsilon_(0))` where dielectric is present `= (Q)/(KA epsilon_(0))`
As `K_(1) lt K_(2)` so `E_(1) gt E_(2)`
Hence graph `(c )` correctly depicts the variation of electric field `E` with distance `d`.
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