= Solution
Use a cross-section normal to the sheets' line of intersection: the positive sheet is $x=0$, and the negative sheet is $y=0$. A <Gaussian pillbox> gives a single infinite sheet's <electric field> magnitude $\sigma/(2\epsilon_0)$ on either side. <Superposition> consequently gives, away from the sheets,
$$
\boxed{\mathbf E=\frac{\sigma}{2\epsilon_0}\bigl(\operatorname{sgn}(x)\hat{\mathbf x}-\operatorname{sgn}(y)\hat{\mathbf y}\bigr),\qquad
|\mathbf E|=\frac{\sigma}{\sqrt2\epsilon_0}.}
$$
The <electric field> points away from the positively charged sheet and towards the negatively charged sheet, bisecting each right angle. The <electric field lines> are straight within each quadrant: in the first quadrant, for example, $dy/dx=-1$, so $x+y$ is constant along a line. They start on $x=0$ and finish on $y=0$. At a sheet, the normal component has the usual surface-charge jump, so the quadrant formula specifies the appropriate one-sided limits rather than a unique <electric field> on the charged surface.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-crossed-charge-sheets.png]
{title=Cross-section of the electric field between perpendicular oppositely charged sheets}
{description=Each arrow starts on the positive sheet and terminates on the negative sheet; the diagram is an original schematic of the derived field.}
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