Final answer:
The charge on the top plate of the parallel-plate capacitor is 5.698 x 10^-11 C when connected to a 6 V battery, having plate area 3976 cm² and separated by 0.37 cm.
Step-by-step explanation:
The question relates to the charge (Q) on the top plate of a parallel-plate capacitor that is connected to a battery with a voltage (Vb) of 6 V. The area (A) of each plate is 3976 cm², and the separation distance (d) between the plates is 0.37 cm. To find the charge on the top plate, we need to use the formula for the capacitance (C) of a parallel-plate capacitor, C = ε0 · (A/d), and then use the formula Q = C · V to get the charge, where ε0 is the vacuum permittivity (8.85 x 10-12 C²/N·m²). Since the question doesn't provide a dielectric constant, we assume that the space between the plates is a vacuum or air which has a dielectric constant close to 1.
First, convert the area from cm² to m² (1 cm² = 1 x 10-4 m²): A = 3976 cm² × 1 x 10-4 m²/cm² = 0.3976 m². Next, convert the separation distance from cm to m (1 cm = 0.01 m): d = 0.37 cm × 0.01 m/cm = 0.0037 m. The capacitance is then calculated as C = (8.85 x 10-12 C²/N·m²) · (0.3976 m² / 0.0037 m) = 9.497 x 10-12 F. Using the capacitance value, we can now calculate the charge: Q = C · V = 9.497 x 10-12 F · 6 V = 5.698 x 10-11 C.
Therefore, the charge on the top plate is 5.698 x 10-11 C.