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(A) Let the square have vertices $A, B, C,$ and $D$ with identical charges $q$ placed at each corner. Let the center of the square be $O$. The distanc

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(A) Let the square have vertices $A, B, C,$ and $D$ with identical charges $q$ placed at each corner. Let the center of the square be $O$. The distance from each corner to the center $O$ is the same,say $r$. The electric field intensity at the center $O$ due to a charge at corner $A$ is $\vec{E}_A$,directed away from $A$ along the diagonal $AC$. Similarly,the electric field intensity due to the charge at the opposite corner $C$ is $\vec{E}_C$,directed away from $C$ along the diagonal $AC$. Since the charges are identical and the distances are equal,the magnitudes are equal: $|\vec{E}_A| = |\vec{E}_C| = E$. Because they act in opposite directions,$\vec{E}_A + \vec{E}_C = 0$. Similarly,for the charges at corners $B$ and $D$,the electric fields $\vec{E}_B$ and $\vec{E}_D$ are equal in magnitude and opposite in direction,so $\vec{E}_B + \vec{E}_D = 0$. The resultant electric field at the center is $\vec{E}_{net} = \vec{E}_A + \vec{E}_B + \vec{E}_C + \vec{E}_D = 0$.

Four identical charges are placed at the four corners of a square. If the electric field intensity at the center of the square due to any one charge is $E$,then the resultant electric field intensity at the center of the square will be .......

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