Solution

The electric field intensity for a single infinite plane with surface charge density $\sigma$ is:

$$E = \frac{\sigma}{2\epsilon_0}$$

Region 1

In the $y$-direction:
$$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\epsilon_0}(1-\cos{\alpha})$$

In the $x$-direction:
$$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$

Using $E_{R1} = \sqrt{E_{R1x}^2 + E_{R1y}^2}$ and the identity $\sin{\frac{\alpha}{2}} = \sqrt{\frac{1-\cos{\alpha}}{2}}$:

**Answer 1:**
$$E_{R1} = \frac{\sigma}{\epsilon_0}\sin{\frac{\alpha}{2}}$$

Region 2

In the $y$-direction:
$$E_{R2y} = E(1+\cos{\alpha}) = \frac{\sigma}{2\epsilon_0}(1+\cos{\alpha})$$

In the $x$-direction:
$$E_{R2x} = E \sin{\alpha} = \frac{\sigma}{2\epsilon_0}\sin{\alpha}$$

Using $E_{R2} = \sqrt{E_{R2x}^2 + E_{R2y}^2}$ and the identity $\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$:

**Answer 2:**
$$E_{R2} = \frac{\sigma}{\epsilon_0}\cos{\frac{\alpha}{2}}$$