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Question

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

Vedclass · Accepted Answer

$\vec E$ changes,$V$ remains unchanged

Vedclass · Answer

(A) The electric potential $V$ at the centre of the square is the algebraic sum of the potentials due to individual charges: $V = \sum \frac{kq}{r}$. 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 the charges does not change the sum $(q_A + q_B + q_C + q_D)$,so $V$ remains unchanged.The electric field $\vec E$ is a vector quantity. The resultant electric field at the centre is the vector sum of the fields due to individual charges. Initially,the charges are $q$ at $A, B$ and $-q$ at $C, D$. After interchanging $A$ with $D$ and $B$ with $C$,the charges at $A$ and $B$ become $-q$,and at $C$ and $D$ become $q$. This effectively reverses the direction of the resultant electric field vector at the centre. Thus,$\vec E$ changes.

Charges are placed on the vertices of a square as shown. Let $\vec E$ be the electric field and $V$ be the 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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