Animated Solution for Physics - Electrostatics: A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x=0$ and positive plate is at $x=3d$. The slab is equidistant from the plates. The capacitor is given some charge. As $x$ goes from $0$ to $3d$

Right in the middle of this setup, spanning from $x=d$ to $x=2d$, we place a dielectric slab of thickness $d$. This creates three distinct regions: an air gap, a dielectric medium, and another air gap.

We know from fundamental electrostatics that electric field lines always originate from positive charges and terminate at negative charges. Since our positive plate is at $x=3d$ and the negative plate is at $x=0$, the electric field must point from right to left. In vector terms, it points in the $-x$ direction everywhere between the plates.

What about its magnitude? In the air gaps ($0<x<d$ and $2d<x<3d$), the electric field has a constant magnitude, let's call it $E_0$. However, inside the dielectric slab ($d<x<2d$), the material gets polarized. This polarization creates an internal electric field that opposes the external one, effectively weakening the net electric field to $E_0/K$, where $K$ is the dielectric constant.

So, while the magnitude of the electric field drops inside the dielectric, its direction remains strictly unchanged. This confirms that option (b) is correct and option (a) is incorrect.

The relationship between the electric field and the potential is given by the gradient equation:

Rearranging this, we get:

Since we established that the electric field $E$ is negative everywhere (pointing in the $-x$ direction), the term $-E$ is always positive. Therefore, for any forward step $dx>0$, the change in potential $dV$ is strictly positive.

Physically, this makes perfect sense: as you move from $x=0$ to $x=3d$, you are walking from the negative plate (lower potential) towards the positive plate (higher potential). You are moving against the electric field lines, which means you are climbing up the potential hill. The potential increases continuously across all three regions. The only difference is that the rate of increase (the slope of the $V-x$ graph) is slightly less steep inside the dielectric because the electric field is weaker there.

Furthermore, traversing from the negative to the positive plate guarantees a continuous increase in electric potential. Thus, the correct statements are (b) and (c).