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(D) Let $d$ be the distance from any vertex to the center of the square.$1$. Electric Potential $(V)$: The potential at the center is the algebraic su

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$\vec{E}$ changes and $V$ remains unchanged.

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(D) Let $d$ be the distance from any vertex to the center of the square.$1$. Electric Potential $(V)$: The potential at the center is the algebraic sum of potentials due to individual charges: $V = \sum \frac{kq_i}{d} = \frac{k}{d} (q_A + q_B + q_C + q_D)$.Initially,$V = \frac{k}{d} (q + q - q - q) = 0$.After interchanging $A$ with $D$ and $B$ with $C$,the new charges at the vertices are $q_A' = -q$,$q_B' = -q$,$q_C' = q$,and $q_D' = q$. The new potential $V' = \frac{k}{d} (-q - q + q + q) = 0$.Thus,$V$ remains unchanged.$2$. Electric Field $(\vec{E})$: The electric field at the center is the vector sum of fields due to individual charges. Initially,the fields due to $A$ and $C$ are in the same direction (away from $A$ towards $C$),and fields due to $B$ and $D$ are in the same direction (away from $B$ towards $D$).After interchanging the charges,the magnitudes of the individual field vectors remain the same,but their directions change because the positions of the charges have been swapped. Specifically,the net field vector $\vec{E}$ rotates by $90^\circ$ or changes direction,so $\vec{E}$ changes.

Charges are placed on the vertices of a square as shown in the figure. Let $\vec{E}$ be the electric field and $V$ be the electric potential at the center. If the charges at $A$ and $B$ are interchanged with those at $D$ and $C$ respectively,then:

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