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Two point charges $+q$ and $-q$ are held fixed at $(-d, 0)$ and $(d, 0)$ respectively of a $x -y$ coordinate system. Then
the electric field $E$ at all points on the axis has the same direction
work has to be done in bringing a test charge from $\infty $ to the orgin
electric field at all points on $y-$ axis is along $x-$ axis
the dipole moment is $2qd$ along the $x-$ axis
Solution
If we take a point $M$ on the $X-$ axis as shown in the figure, then the net electric field is in $X-$ direction.
$\therefore $ Option $(a)$ is incorrect.
If we take a point $N$ on $Y-$ axis, we find net electric field along $+X$ direction. The same will be true for any point on $Y-$ axis. $(c)$ is a correct option.
${W_{ \propto 0}} = q({V_ \propto } – {V_0}) = q(0 – 0) = 0$
$\therefore $ $(b)$ is incorrect. The direction of dipole moment is from $-ve$ to $+ve$. Therefore $(d)$ is incorrect.
