Question

Two infinite planes each with uniform surface charge density $$+\sigma$$ are kept in such a way that the angle between them is $$30^0$$. The electric field in the region shown between them is given by :

Answer 1

Correct option is A. $$\dfrac{\sigma}{2\epsilon_0}\left[\left(1-\dfrac{\sqrt{3}}{2}\right)\hat{y}-\dfrac{{\hat{x}}}{2}\right]$$
Electrical field due to an infinite sheet is given by $$\dfrac{\sigma}{2\epsilon_0}$$

where $$\sigma$$ is the surface charge density.

$$\vec{E_1}$$ is the electric field due to the first sheet and can be given as:

$$\vec{E_1}=\dfrac{\sigma}{2\epsilon_0}[-\cos 60^0\hat{x}-\sin 60^0\hat{y}]$$

$$\vec{E_2}$$ is the electric field due to the second sheet and can be given as:

$$\vec{E_2}=\dfrac{\sigma}{2\epsilon_0}\hat{y}$$

$$\vec{E_1}+\vec{E_2}=\left(-\dfrac{\sigma}{4\epsilon_0}\hat{x}-\dfrac{\sqrt{3}\sigma}{4\epsilon_0}\hat{y}\right)+\dfrac{\sigma}{2\epsilon_0}\hat{y}$$

$$=\dfrac{\sigma}{2\epsilon_0}\left[(1-\dfrac{\sqrt{3}}{2})\hat{y}-\dfrac{\hat{x}}{2}\right]$$

So, $$option(A)$$ is correct.