Key idea about the center of a square

Let the square have vertices $A,B,C,D$ and center $O$.

The distance of every vertex from the center is the same. Let this distance be $r$.

Therefore, the potential at the center is

Effect on potential when charges are interchanged

We are told that the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively.

That means the set of charges at the four corners is the same; only their positions change.

Hence, $q_A+q_B+q_C+q_D$ remains unchanged.

Therefore, $V \text{ remains unchanged.}$

Now consider the electric field at the center

Electric field is a vector quantity, so arrangement matters.

The field at the center due to a charge at a vertex is along the line joining that vertex to the center. Since opposite vertices correspond to opposite directions, changing which charge sits at which corner generally changes vector addition.

Vector form argument

Let the position vectors of vertices relative to the center be: $\vec r_A,\vec r_B,\vec r_C,\vec r_D$ with $\vec r_C=-\vec r_A, \qquad \vec r_D=-\vec r_B.$

Since all vertices are at the same distance $r$ from the center, the electric field at the center is proportional to $\vec E \propto -(q_A\vec r_A+q_B\vec r_B+q_C\vec r_C+q_D\vec r_D).$

Using opposite-vertex relations, $\vec E \propto -\big[(q_A-q_C)\vec r_A + (q_B-q_D)\vec r_B\big].$

After interchange:

• charge at $A$ becomes old charge at $D$,
• charge at $B$ becomes old charge at $C$,
• charge at $C$ becomes old charge at $B$,
• charge at $D$ becomes old charge at $A$.

So new field is $\vec E' \propto -\big[(q_D-q_B)\vec r_A + (q_C-q_A)\vec r_B\big].$

In general, this is not equal to $\vec E$.

Hence, $\vec E \text{ changes.}$

Conclusion

• Electric potential at the center remains unchanged.
• Electric field at the center changes.

Therefore, the correct option is $\boxed{\text{A: } \vec E \text{ changes, } V \text{ remains unchanged}}.$