The scenario involves three large parallel conducting charged sheets, commonly denoted as A, B, and C. To determine the net force between the charged plates A and B, we must analyze the electric fields created by each sheet.
Electric Field Due to a Single Sheet
- Each large conducting sheet generates an electric field that is uniform and perpendicular to its surface.
- The electric field (E) due to an infinite sheet of charge with surface charge density σ is given by:
E = σ / (2ε₀)
where ε₀ is the permittivity of free space.
Electric Fields Between the Sheets
- For three sheets A, B, and C, assume:
- Sheet A has charge density +σ
- Sheet B has charge density -σ
- Sheet C has charge density +σ
- The electric fields in the regions between the sheets are:
- Between A and B:
- E_AB = E_A + E_B = (σ / (2ε₀)) + (σ / (2ε₀)) = σ / ε₀ (directed from A to B)
- Between B and C:
- E_BC = E_B + E_C = (−σ / (2ε₀)) + (σ / (2ε₀)) = 0 (no net electric field)
Net Force Between Plates A and B
- The net force (F) between the plates is determined by the electric field acting on the charge of the plates.
- The force on sheet A due to the electric field created by sheet B is given by:
F = qE
where q is the charge on sheet A, and E is the electric field between the sheets.
- The electric field E_AB = σ / ε₀, hence:
F_AB = q(σ / ε₀)
- Similarly, the force on sheet B due to the electric field from sheet A is equal in magnitude and opposite in direction:
F_BA = -q(σ / ε₀)
Thus, the net force between plates A and B is attractive due to opposite charges, leading to a mutual attraction.