Final answer:
The electric field strength between the plates of a 0.90-μF air-gap capacitor with a 1.0 mm separation and a charge of 73 μC on each plate is 8.1 × 10⁴ N/C.
Step-by-step explanation:
To determine how strong the electric field is between the plates of a 0.90-μF air-gap capacitor with 73 μC of charge on each plate and a separation of 1.0 mm, we can use the formula for the electric field (E) between the plates of a parallel plate capacitor, E = Q / (ε0A), where Q is the charge on the plates, ε0 is the permittivity of free space, and A is the area of the plates.
First, we need to find the area (A) of the plates using the capacitance (C) formula, C = ε0A / d, where d is the separation between plates. Rearranging for A gives A = C × d / ε0. The values are C = 0.90 × 10-6 F, d = 1.0 × 10-3 m, and ε0 = 8.85 × 10-12 C2/N·m2. Using these, we can solve for A.
After finding A, we can insert this value back into the formula for E. Based on the student's available options, we are only looking for the magnitude of the electric field.
If every step of the calculation is done correctly, the magnitude of the electric field (E) is found to be one of the options provided, 8.1 × 104 N/C (Option B), representing the strength of the electric field between the plates of the given capacitor.