Two infinite planes each with uniform surface charge density $$+\sigma$$ are kept in such a way that the angle between them is 30$$^\circ$$. The electric field in the region shown between them is given by:

$$\text{The electric field in the region above it points straight up along the positive y-axis:}$$

$$\vec{E}_1 = \frac{\sigma}{2\varepsilon_0}\hat{y}$$

For the inclined sheet (inclined at $$30^\circ$$ to the horizontal with charge density $$+\sigma$$):

$$\text{The normal to this sheet points into the region between them, directed downwards and leftwards.}$$

$$\text{The angle of this electric field vector with the negative y-axis is } 30^\circ$$ making a $$60^\circ$$ angle with the negative $$x$$-axis:

$$\vec{E}_2 = E \cos(60^\circ)(-\hat{x}) + E \sin(60^\circ)(-\hat{y})$$

$$\vec{E}_2 = \frac{\sigma}{2\varepsilon_0}\left(-\frac{1}{2}\hat{x} - \frac{\sqrt{3}}{2}\hat{y}\right)$$

For the net electric field $$\vec{E}_{\text{net}}$$ in the region between them:

$$\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2$$

$$\vec{E}_{\text{net}} = \frac{\sigma}{2\varepsilon_0}\hat{y} + \frac{\sigma}{2\varepsilon_0}\left(-\frac{1}{2}\hat{x} - \frac{\sqrt{3}}{2}\hat{y}\right)$$

$$\vec{E}_{\text{net}} = \frac{\sigma}{2\varepsilon_0}\left[\left(1 - \frac{\sqrt{3}}{2}\right)\hat{y} - \frac{1}{2}\hat{x}\right]$$