Electric field due to negatively charged plate towards that plate and is equal to sigma/ 2ephslanot.electic field due to positively charged plate is away from it and is equal to Sigma/2 ephslano. since both are in same direction they are added and we get option 'b'as answer

$\sigma_1$ and $\sigma_2$ respectively are separated by a distance $d$. The electric field between the plates can be found using Gauss's law.

Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.

In this case, we can choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the plates. The electric field between the plates is uniform and perpendicular to the surface of the plates, so the electric flux through the curved surface of the cylinder is zero.

The electric flux through the top and bottom surfaces of the cylinder is also zero because the electric field is parallel to these surfaces. Therefore, the only contribution to the electric flux comes from the two flat surfaces of the cylinder which are perpendicular to the electric field.

The electric flux through each of these surfaces is given by:
$\Phi = E \cdot A$

where $E$ is the electric field and $A$ is the area of each surface.

The area of each surface is equal to the product of the length of the cylinder ($L$) and the distance between the plates ($d$).

Therefore, the electric flux through each surface is:
$\Phi = E \cdot L \cdot d$

Since the electric field is uniform between the plates, the electric field is the same at each point on the plates. Therefore, the electric field can be taken out of the integral sign.

The total electric flux through the Gaussian surface is the sum of the fluxes through each surface. Therefore, the total electric flux is:
$\Phi = 2 E \cdot L \cdot d$

According to Gauss's law, the electric flux is also equal to the total charge enclosed divided by the permittivity of the medium. Therefore, we have:
$\Phi = \frac{Q_{enc}}{\epsilon_0}$

where $Q_{enc}$ is the charge enclosed by the Gaussian surface and $\epsilon_0$ is the permittivity of free space.

Since the plates are infinitely long, the charge enclosed by the Gaussian surface is equal to the surface charge density times the area of the Gaussian surface. Therefore, we have:
$\Phi = \frac{\sigma_1 A + \sigma_2 A}{\epsilon_0}$

Substituting the expression for the area of the Gaussian surface, we get:
$\Phi = \frac{\sigma_1 L d + \sigma_2 L d}{\epsilon_0}$

Simplifying the expression, we get:
$\Phi = \frac{(\sigma_1 + \sigma_2) L d}{\epsilon_0}$

Setting this equal to the expression we obtained earlier for the electric flux, we get:
$2 E \cdot L \cdot d = \frac{(\sigma_1 + \sigma_2) L d}{\epsilon_0}$

Simplifying the expression, we get:
$2 E = \frac{(\sigma_1 + \sigma_2)}{\epsilon_0}$

Therefore, the electric field between the plates is:
$E = \frac{\sigma_1 + \sigma_2}{2 \epsilon_0}$