Two dielectric slabs of area of cross-section same as the area of the plates are introduced inside a capacitor as shown. Now, the capacitor is charged. If the potential of the upper plate of the capacitor is
`V_(H)` and the potential of the lower plate is `V_(L)`, the potential at the interface of the two slabs is:

Options

A. `(1)/(3)(V_(H)-2V_(L))`
B. `(1)/(3)(2V_(H)-V_(L))`
C. `(1)/(3)(V_(H)-V_(L))`
D. `(2)/(3)(V_(H)-V_(L))`

Answer

A

Explanation

Let us assume that `V_(H) gt V_(L)`. This assumption will not affect our answer, and you can test this yourself by taking the opposite assumption and solving.
Let the electric field insie the upper slab be `E_(H)` and the electric inside the lower slab be `E_(L)`.
Then, we know that `(E_(H))/(E_(L))=(K_(2))/(K_(1))=2`
Let the potential at the interface be V.
Then, `V_(H)-V=E_(H)((d)/(2)) and V-V_(L)=E_(L)((d)/(2))`
Solving, we get `V=(1)/(3)(V_(H)+2V_(L))`