Figure shows portions of two large, parallel, non-conducting sheets, each with a fixed uniform charge on one side. The magnitudes of the surface charge densities are α=6.8+X μC/m2 for the positively charged sheet and α=4.3+Y μC/m2 for the negatively charged sheet.
Find the electric field (a) to the left of the sheets (b) between the sheets, and (c) to the right of the sheets.
X=3 y=6 z=5
To find the electric field in the given scenario, we can make use of the principle of superposition. According to this principle, the total electric field at a certain point due to multiple charges or charge distributions is equal to the vector sum of the electric fields produced by each individual charge or distribution.
(a) To find the electric field to the left of the sheets:
Since there is only the positively charged sheet to the left of the point, the electric field at that point will only be due to the positively charged sheet. The electric field E1 produced by a uniformly charged sheet is given by the equation E1 = σ / (2ε0), where σ is the surface charge density and ε0 is the permittivity of free space. In this case, the surface charge density is α = 6.8 + X μC/m². Therefore, the electric field to the left of the sheets is E1 = (6.8 + X) / (2ε0).
(b) To find the electric field between the sheets:
The electric field between the sheets will be the vector sum of the electric fields produced by both sheets. The electric field due to the positively charged sheet, E1, will be in the opposite direction to the electric field due to the negatively charged sheet, E2. Therefore, the net electric field magnitude between the sheets will be the difference between |E1| and |E2|. Since the surface charge density of the negatively charged sheet is α = 4.3 + Y μC/m², the electric field due to the negatively charged sheet is E2 = (4.3 + Y) / (2ε0). Hence, the electric field between the sheets is |E1 - E2| = |(6.8 + X) / (2ε0) - (4.3 + Y) / (2ε0)|.
(c) To find the electric field to the right of the sheets:
Similar to part (a), the electric field to the right of the sheets will only be due to the negatively charged sheet. Therefore, the electric field to the right of the sheets is E2 = (4.3 + Y) / (2ε0).
Note: Remember that electric field is a vector quantity, so it has both magnitude and direction.