# Problem with the formula for electric intensity due to a charged sheet

The formula for the electric field due to a charged sheet is $$\LARGE\frac{\sigma}{2\epsilon_0}$$. Now, if I want to find the electric intensity of a point between two equally and oppositely charged plates, with the same charge densities, I would have to add the two electric intensities: $$\Large\frac{\sigma}{2\epsilon_0} + \Large\frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$$

My problem with this is, why aren't we taking the charge density as negative for the negatively charged plate? If we were to do it that way, it would look something like this: $$\Large\frac{\sigma}{2\epsilon_0} - \Large\frac{\sigma}{2\epsilon_0} = 0$$

Isn't it against the rules of physics to take the charge of a negatively charged plate as positive? To elaborate:

Charge on the plate = $$-q$$

Area of the plate = $$-A$$

So, charge density = $$-\frac{q}{A}$$

$$\implies$$ charge density = $$-\sigma$$

Clearly, we can see that the charge density of a negatively charged plate is negative. Why are we doing otherwise in this case?

Actually, we indeed take the minus sign. But, the direction of electric field due to the negative plate will be in the opposite direction of that of positive plate. Mathematically understanding, we have \begin{align}\vec E&=\vec E_1+\vec E_2\\ &=\frac{\sigma}{2\epsilon_0}\ \hat i+\frac{(-\sigma)}{2\epsilon_0}\ (-\hat i) \\ &=\frac{\sigma}{\epsilon_0}\ \hat i\end{align}