= Solution
Use $x=\rho\cos\phi$, $y=\rho\sin\phi$ to recognize the <planar quadrupole potential>, $V=A(x^2-y^2)$. Its <electric field> is minus the <gradient>:
$$
\boxed{\mathbf E=(-2Ax,\,2Ay,\,0),}
$$
or, in <cylindrical coordinates>,
$$
E_\rho=-2A\rho\cos2\phi,\qquad
E_\phi=2A\rho\sin2\phi,\qquad E_z=0.
$$
For $A>0$, arrows point toward the origin along the $x$ axis and away from it along the $y$ axis. The off-axis <electric field lines> satisfy
$$
\frac{dy}{dx}=\frac{E_y}{E_x}=-\frac yx,
\qquad \boxed{xy=\mathrm{constant}.}
$$
The axes themselves are the separating field lines; the origin is a field null. Reversing $A$ reverses every arrow. The field is independent of $z$, and $A$ has units $\mathrm{V\,m^{-2}}$.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2017/paper-1-a3-quadrupole-field.png]
{title=Original calculated electric-field lines and equipotentials for the planar quadrupole with positive A; arrows approach along x and leave along y}
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