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(D) The electric potential $V$ at the centre of the square is the algebraic sum of potentials due to individual charges: $V = k \sum \frac{q_i}{r}$. S

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$\overrightarrow{E}$ changes,$V$ remains unchanged

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(D) The electric potential $V$ at the centre of the square is the algebraic sum of potentials due to individual charges: $V = k \sum \frac{q_i}{r}$. Since the distance $r$ from each vertex to the centre is the same,$V = \frac{k}{r} (q_A + q_B + q_C + q_D)$. Interchanging charges does not change the sum $(q_A + q_B + q_C + q_D)$,so $V$ remains unchanged.The electric field $\overrightarrow{E}$ at the centre is a vector sum: $\overrightarrow{E} = \sum \overrightarrow{E_i}$. Initially,the field vectors from $A$ and $C$ (both $q$) and $B$ and $D$ (both $-q$) result in a net field. When charges are interchanged,the positions of the charges change relative to the centre,which alters the direction and magnitude of the individual field vectors $\overrightarrow{E_i}$. Thus,the resultant vector $\overrightarrow{E}$ changes.

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

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