For the given figure,the direction of the ele | vedclass

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(B) The electric field at point $A$ due to the positive charge $+Q$ at point $B$ is directed away from $B$,which is along the direction $AX$.The elect

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towards $AY$

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(B) The electric field at point $A$ due to the positive charge $+Q$ at point $B$ is directed away from $B$,which is along the direction $AX$.The electric field at point $A$ due to the negative charge $-Q$ at point $C$ is directed towards $C$,which is along the direction $AZ$.Since the magnitudes of the charges are equal and point $A$ is equidistant from $B$ and $C$,the magnitudes of the electric fields $E_{AB}$ and $E_{AC}$ are equal.The resultant electric field is the vector sum of $E_{AB}$ and $E_{AC}$. According to the parallelogram law of vector addition,the resultant vector of two equal vectors lies along the angle bisector of the angle between them.In the given figure,the direction $AY$ represents the angle bisector of the angle between $AX$ and $AZ$. Therefore,the resultant electric field at $A$ is in the direction of $AY$.

For the given figure,the direction of the electric field at point $A$ will be:

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