Answer :
To solve this problem, we need to understand the role of the electric field ([tex]E_d[/tex]) between two charged plates and how it relates to the surface charge densities on the plates. The surface charge densities are represented as [tex]\sigma_a[/tex], [tex]\sigma_b[/tex], [tex]\sigma_c[/tex], and [tex]\sigma_d[/tex] for each surface of the plates involved.
The relationship between the electric field and surface charge density in a parallel plate capacitor can be expressed using Gauss's Law. According to Gauss's Law, the electric field [tex]E[/tex] between the plates of a capacitor depends on the surface charge density [tex]\sigma[/tex] and the permittivity of free space [tex]\varepsilon_0[/tex]. The equation can be represented as:
[tex]E = \frac{\sigma}{\varepsilon_0}[/tex]
For two plates with charge densities, the electric field due to a single surface is [tex]\sigma/\varepsilon_0[/tex], and considering two plates, we would accommodate the vectors jointly.
In the given choices, the relationship phrased as [tex]\sigma_a - \sigma_b - \sigma_c + \sigma_d = 2\varepsilon_0 E_d[/tex] most effectively resembles the expression derived from Gauss's Law considering systems with multiple surfaces impacting the field. When you combine the contributions of each surface and account for their orientation and distance, this relation should correctly connect the electric field and charge densities across the plates.
Thus, the correct option is:
- [tex]\sigma_a - \sigma_b - \sigma_c + \sigma_d = 2\varepsilon_0 E_d[/tex]
This expression gives an appropriate representation of how the charge densities on the surfaces of the plates contribute to the electric field magnitude in this setup.
