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(A) The system forms an electric dipole with charges $+q$ at $(-d, 0)$ and $-q$ at $(+d, 0)$.$1$. For points on the $y$-axis,the electric field due to

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$E$ at all points on the $y$-axis is along $-\hat{i}$

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(A) The system forms an electric dipole with charges $+q$ at $(-d, 0)$ and $-q$ at $(+d, 0)$.$1$. For points on the $y$-axis,the electric field due to the dipole is directed from the positive charge to the negative charge,which is in the $-\hat{i}$ direction.$2$. For points on the $x$-axis,the direction of the electric field changes depending on the position relative to the charges.$3$. The dipole moment $\vec{p}$ is directed from $-q$ to $+q$,which is from $(+d, 0)$ to $(-d, 0)$,i.e.,in the $-\hat{i}$ direction. Its magnitude is $p = q(2d) = 2qd$.$4$. The electric potential $V$ at the origin $(0, 0)$ is $V = \frac{kq}{d} + \frac{k(-q)}{d} = 0$. Since the potential at the origin is zero,the work done to bring a test charge from infinity to the origin is $W = q_0 \Delta V = q_0(0 - 0) = 0$.

Two point charges $+q$ and $-q$ are held fixed at $(-d, 0)$ and $(+d, 0)$ respectively of an $(x, y)$ coordinate system. Then:

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