Final answer:
The charge on the positive electrode of the parallel-plate capacitor with the given dimensions and electric field is 3.54 nC.
Step-by-step explanation:
To calculate the charge on the positive electrode of a parallel-plate capacitor, we can use the relationship between electric field (E), charge (Q), and the area of the plates (A). The electric field in a capacitor is given by E = Q / (ε₀A), where ε₀ is the permittivity of free space (ε₀ = 8.85 \u00d7 10^-12 C^2/Nm^2), and A is the area of the plates.
The given values are: E = 1.0 \u00d7 10^6 N/C, and the plate area is 2.0 cm \u00d7 2.0 cm. Note that the area should be converted to square meters (m^2) for consistency in units: A = (2.0 \u00d7 10^-2 m) \u00d7 (2.0 \u00d7 10^-2 m) = 4.0 \u00d7 10^-4 m^2.
Using the formula, we can solve for Q:
Q = E \u00d7 ε₀ \u00d7 A = (1.0 \u00d7 10^6 N/C) \u00d7 (8.85 \u00d7 10^-12 C^2/Nm^2) \u00d7 (4.0 \u00d7 10^-4 m^2) = 3.54 \u00d7 10^-9 C.
Since 1 C = 10^9 nC, the charge Q in nanocoulombs (nC) is Q = 3.54 nC.