= Solution
An infinite sheet of <surface charge density> $\sigma$ gives an <electric field> of magnitude $E_s=\sigma/(2\varepsilon_0)$ on either side, normal to the sheet and directed away from positive charge. Use the <superposition principle> for the two sheets. Choose normals $\widehat{\mathbf y}$ for $y=0$ and $\mathbf n=(-\widehat{\mathbf x}+\widehat{\mathbf y})/\sqrt2$ for $y=x$. Away from the charged planes,
$$
\boxed{\mathbf E=E_s\left[\operatorname{sgn}(y)\widehat{\mathbf y}
-\operatorname{sgn}(y-x)\frac{-\widehat{\mathbf x}+\widehat{\mathbf y}}{\sqrt2}\right].}
$$
The minus sign on the second contribution makes its field point toward the negatively charged sheet. Each sector has a uniform <electric field>; its direction changes on crossing a sheet.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1-a1-crossed-charge-sheets.png]
{title=Original calculated electric-field directions for the intersecting positive and negative charge sheets; arrows retain the relative field magnitudes in the four sectors}
For example, above both sheets the field points right and slightly upward, whereas in the sector $y>0$ but $y<x$ it points left and steeply upward. The other two sectors follow by inversion through the origin. The normal field jumps by the appropriate surface charge divided by $\varepsilon_0$; a single field value on an ideal zero-thickness charged sheet is not required.
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