Electric field between widening pair of conductors

The question is: There are two plates of conductors as shown in figure below. Among the four possibilities shown in figure, which one has the electric field lines correctly? 

Here for the answer:

Firstly let us reduce the possibility. In a conductor, there can never be any tangent electric field close to surface of conductor.  So the electric field going in/out must be perpendicular to the surface clost to the conductor. Hence (a) and (b) are not possible.

In (c) and (d), we see that in (c) the electric field is denser in the narrower edge than wider edge of the conductor. (d) is uniformly spread electric field. To make this choice, we play smart, and have the following setup: two capacitor of same area but the distance between the plates are narrower and the other wider. They are connected to wire. Doesn't it look similar to the setup in question, just that we can solve for this.

We can see that the capacitance of the narrower sepration is more than the wider sepration. i.e., \( C_{narrower} > C_{wider} \).

Now we use \( Q=CV \) to figure out which contain more charges. Since they both are at the same potential (as they are connected by the wire), we have:

$$ V=Q_{narrow}/C_{narrow}=Q_{wide}/C_{wide} $$

Since \( C_{narrow} > C_{wide} \) , \( \therefore Q_{narrow} > Q_{wide} \) .  And more Q means more electric field. Hence choice (c) is correct.

THE END