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(B) Let the charges at vertices $A$,$B$,and $C$ be $q$. The mid-point $M$ of the hypotenuse $BC$ is equidistant from $B$ and $C$. Let $\vec{E_A}$,$\ve

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$2$

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(B) Let the charges at vertices $A$,$B$,and $C$ be $q$. The mid-point $M$ of the hypotenuse $BC$ is equidistant from $B$ and $C$. Let $\vec{E_A}$,$\vec{E_B}$,and $\vec{E_C}$ be the electric fields at $M$ due to charges at $A$,$B$,and $C$ respectively. The electric field $\vec{E_B}$ due to charge at $B$ is directed away from $B$ along $BM$,and $\vec{E_C}$ due to charge at $C$ is directed away from $C$ along $MC$. Since $MB = MC$ and the charges are identical,$|\vec{E_B}| = |\vec{E_C}|$. These two vectors are equal in magnitude and opposite in direction,so they cancel each other out $(\vec{E_B} + \vec{E_C} = 0)$. The net electric field at $M$ is therefore $\vec{E_{net}} = \vec{E_A} + \vec{E_B} + \vec{E_C} = \vec{E_A}$. The field $\vec{E_A}$ is directed away from the charge at $A$ towards $M$. Looking at the provided vector diagram,this direction coincides with vector $2$.

Three identical point charges are placed at the vertices of an isosceles right-angled triangle as shown. Which of the numbered vectors coincides in direction with the electric field at the mid-point $M$ of the hypotenuse?

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