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(B) The electric field due to a negative charge is directed towards the charge.Let the two negative charges be $-Q$ located at $(0, a)$ and $(0, -a)$.

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along negative $x$-direction

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(B) The electric field due to a negative charge is directed towards the charge.Let the two negative charges be $-Q$ located at $(0, a)$ and $(0, -a)$. Point $P$ is on the positive $x$-axis at $(x, 0)$.The electric field vector $\vec{E}_1$ due to the charge at $(0, a)$ points from $P$ towards $(0, a)$.The electric field vector $\vec{E}_2$ due to the charge at $(0, -a)$ points from $P$ towards $(0, -a)$.Since the magnitudes of the charges are equal and their distances from $P$ are equal,the magnitudes of the electric fields are equal,i.e.,$|\vec{E}_1| = |\vec{E}_2|$.When resolving these vectors into components,the $y$-components ($E_{1y}$ and $E_{2y}$) are equal in magnitude but opposite in direction,so they cancel each other out.The $x$-components ($E_{1x}$ and $E_{2x}$) both point in the negative $x$-direction.Therefore,the resultant electric field at point $P$ is directed along the negative $x$-direction.

Two negative charges of equal magnitude are located in the $xy$-plane as shown in the figure. The direction of the electric field at point $P$ is

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