The distance between plates of a parallel plate capacitor is `5d`. Let the positively charged plate is at `x=0` and negatively charged plate is at `x=5d`. Two slabs one fo conductor and other of a dielectric of equal thickness `d` are inserted between the plates as shown in figure. Potential verus distance graph will look like:

Options

A.
B.
C.
D.

Answer

2

Explanation

`V_(1)-V_(2)=kq((1)/(r_(1))-(1)/(r_(2)))`
`r_(2)-r_(1)=((V_(1)-V_(2))r_(1)r_(2))/(kq)`, but (r_(2)-r_(1))=t`
`:.ralpha_(1)r_(2)`
If P.D is constant then `(r_(2)-r_(1))=t`