Two thin dielectric slabs of dielectric constants K 1 and K 2 K 1

Question

Two thin dielectric slabs of dielectric constants

K_{1}

and

K_{2} (K_{1} < 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:

Aicharya · Accepted Answer

Here we have to calculate the graph between E and d in the given situation

We know that in dielectric E becomes E/k

where

E = Electric fjeld

k = Dielectric constants

Since $(K_{1} < K_{2})$ , The third option will be correct.

Two thin dielectric slabs of dielectric constants K 1 and K 2 K 1 < K 2 are inserted between plates of a…