Solution
For the net electric field to vanish at point $P(x, x+y)$, the field produced by the point charge at the origin ($\vec{E}_Q$) must be equal and opposite to the field produced by the dipole at $(0,y)$ ($\vec{E}_{dip}$). This implies their vectors must be anti-parallel (or parallel lines of action).
1. Direction of Point Charge Field ($\vec{E}_Q$):
The vector from the origin $(0,0)$ to $P(x, x+y)$ has a slope:
2. Direction of Dipole Field ($\vec{E}_{dip}$):
The dipole is at $(0,y)$ pointing in the $+x$ direction ($\vec{p} = p\hat{i}$). The relative position of point $P$ with respect to the dipole is $(x, x)$. This makes an angle of $45^\circ$ with the dipole axis.
The field of a dipole at a point with position vector $\vec{r}’$ is given by:
$$\vec{E} \propto 3(\vec{p} \cdot \hat{r}’)\hat{r}’ – \vec{p}$$Since the angle is $45^\circ$, $\vec{p}$ and $\hat{r}’$ have a dot product involving $\cos(45^\circ) = 1/\sqrt{2}$.
The vector components resolve such that the resulting field vector has a slope of $3$ (Standard result for $45^\circ$ position: $\tan \alpha = \frac{1}{2} \tan \theta$, where $\alpha$ is angle with radial vector. Or simply, slope $m_{dip} = 3$).
3. Equating Slopes:
For the vectors to cancel, they must lie on the same line:
Answer: (b) 2x