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Understanding surface charge density is crucial when dealing with stationary electric charges spread over a surface. It's defined as the amount of electric charge per unit area. In mathematical terms, it is expressed as \( \sigma = \frac{Q}{A} \), where \( Q \) is the total charge and \( A \) represents the area over which the charge is distributed. With the given problem, the surface charge density is negative for both sheets, indicating an excess of electrons.

These values are not just random numbers; they define the strength of the electric field that the sheets will produce. A higher absolute value of the surface charge density results in a stronger electric field emanating from the surface. Additionally, the negative sign of the surface charge density in our problem suggests that we're dealing with fields that originate from negatively charged surfaces, which is important when determining the direction of the electric fields.

When you hear 'infinite planar charge distribution,' think of a flat sheet with charges that extends forever in all directions, at least in theory. The electric field produced by such a sheet is constant in magnitude and direction at any point near the sheet, provided that you're not too close to the edges — which is why we can treat the sheets in our problem as infinite. The electric field (\( E \)) due to a sheet with a surface charge density (\( \sigma \)) is given by \( E = \frac{\sigma}{2\epsilon_0} \), where \( \epsilon_0 \) is the vacuum permittivity, a constant that characterizes the strength of the electric field in a vacuum.

The beauty of an infinite sheet's electric field is its simplicity. Unlike point charges, which have fields that weaken with distance, an infinite sheet generates a field that is uniform in strength regardless of the distance from the sheet. This makes calculating the electric field at various points especially straightforward in problems like the one we're examining.

To determine the net electric field when more than one field interacts at a point, we use vector addition. Since we're dealing with parallel sheets, we can treat the electric fields as vectors along a single line, simplifying the process to just adding or subtracting scalar quantities depending on their directions.

For instance, in our exercise, a point to the right of sheet A experiences electric fields from both sheets A and B. Since like charges repel, the field produced by sheet A directs to the left while the field due to sheet B directs to the right. Here, the direction comes into play, as one essentially needs to subtract the magnitudes of these fields to find the net field (\( E_{netA} = E_B - E_A \)). Conversely, for a point between the sheets or to the left of sheet A, both fields direct to the left, adding them up (\( E_{netB} = E_A + E_B \)).

It's essential to note that the direction of the field is dependent on the nature of the charge on the sheets. If our sheets were positively charged, the directions would be reversed. The calculation method remains the same, but the vector’s direction would change, a fact that could fundamentally alter the problem’s outcome.