Answer:
A parallel-plate capacitor consists of two parallel conductors with a dielectric material (e.g. plastic) between them. The capacitor is charged by connecting it to a power source, such as a battery. The voltage applied to the plates causes charges of opposite signs to accumulate on the opposing surfaces of the plates.
The charge on each plate is given by: Q = CV where Q is the charge on each plate, C is the capacitance of the capacitor, and V is the voltage applied across the plates.
The capacitance of a parallel-plate capacitor is given by: C = ε0 * A / d where ε0 is the dielectric constant of the material between the plates, A is the surface area of each plate, and d is the distance between the plates.
For the given values: A = (1.5 × 10-4) m2 d = 1.0 mm ε0 = 8.8542 × 10-12 F/m
Plugging these values into the formula for capacitance, we get: C = 8.8542 × 10-12 F/m * (1.5 × 10-4) m2 / 1.0 mm C = 1.32825 × 10-7 F/m
Since the capacitor is connected to a 12-V battery, the voltage applied to the plates is equal to 12 V.
Plugging the values of the capacitance, the voltage, and the plate area into the formula for charge, we get: Q = 1.32825 × 10-7 F/m * 11.999 m2 / 1.0 mm Q = 16.5 × 10-3 C
Therefore, the charge on each plate is 16.5 × 10-3 C or 0.000165 C. If both plates have the same charge, the total charge on the capacitor is twice that value, or 0.00033 C.
Step-by-step explanation: