1. Electric field due to one infinite charged plane

For an infinite plane sheet with uniform surface charge density $\sigma$, the electric field on either side has magnitude

$E=\frac{\sigma}{2\varepsilon_0}$

and is directed normal to the plane, away from the plane if $\sigma>0$.

2. Geometry of the two planes

The two planes make an angle of $30^\circ$ with each other. The electric field in the shown region is the vector sum of the fields due to the two planes.

Let us resolve the two field vectors into $x$ and $y$ components.

Since each field is perpendicular to its plane, the angle between the normals is also $30^\circ$.

From the figure implied by the options, one plane is such that its normal in the region is along $+\hat y$, while the other contributes a field making $30^\circ$ with the negative $x$ side toward positive $y$. Thus:

• Field due to first plane: $\vec E_1=\frac{\sigma}{2\varepsilon_0}\hat y$

• Field due to second plane: its components are $E_{2x}=-\frac{\sigma}{2\varepsilon_0}\sin30^\circ=-\frac{\sigma}{4\varepsilon_0}$ $E_{2y}=\frac{\sigma}{2\varepsilon_0}\cos30^\circ=\frac{\sqrt3\sigma}{4\varepsilon_0}$

So,

$\vec E_2=\frac{\sigma}{2\varepsilon_0}\left(-\frac12\hat x+\frac{\sqrt3}{2}\hat y\right)$

3. Resultant electric field

Now add the two fields:

$\vec E=\vec E_1+\vec E_2$

$\vec E=\frac{\sigma}{2\varepsilon_0}\hat y+\frac{\sigma}{2\varepsilon_0}\left(-\frac12\hat x+\frac{\sqrt3}{2}\hat y\right)$

$\vec E=\frac{\sigma}{2\varepsilon_0}\left[\left(1+\frac{\sqrt3}{2}\right)\hat y-\frac12\hat x\right]$

4. Match with the options

This expression has:

• $y$-component positive,
• $x$-component negative.

Among the options, the one with negative $x$-component is C. But let us check carefully the coefficient of $\hat y$:

Our result:

$\frac{\sigma}{2\varepsilon_0}\left[\left(1+\frac{\sqrt3}{2}\right)\hat y-\frac12\hat x\right]$

Option C is:

$\frac{\sigma}{2\varepsilon_0}\left[(1+\sqrt3)\hat y-\frac12\hat x\right]$

which has an incorrect $y$-coefficient.

Option D is:

$\frac{\sigma}{2\varepsilon_0}\left[\left(1-\frac{\sqrt3}{2}\right)\hat y-\frac12\hat x\right]$

This also does not match our derived result.

So the physically derived expression is

$\boxed{\vec E=\frac{\sigma}{2\varepsilon_0}\left[\left(1+\frac{\sqrt3}{2}\right)\hat y-\frac12\hat x\right]}$

and none of the listed options matches exactly.

5. Comparison with stored answer

Stored correct answer is D, but our derivation shows that D is not correct.

The most likely issue is a typographical error in the options. The correct option should have been:

$\frac{\sigma}{2\varepsilon_0}\left[\left(1+\frac{\sqrt3}{2}\right)\hat y-\frac12\hat x\right]$