Final answer:
The electric field between the electrodes is 40000 V/cm. The capacitance of the capacitor is 2.22 x 10^-10 F. The charge on each electrode is 4.44 x 10^-8 C.
Step-by-step explanation:
To calculate the electric field between the electrodes, we can use the formula E = V/d, where E is the electric field, V is the voltage, and d is the distance between the electrodes. In this case, the voltage is 200 V and the distance is 0.50 mm (or 0.0050 cm). Plugging in these values, we get E = 200 V / 0.0050 cm = 40000 V/cm.
To calculate the capacitance of the capacitor, we can use the formula C = εA/d, where C is the capacitance, ε is the permittivity of the material between the electrodes (for aluminum in air, ε ≈ 8.85 x 10^-12 F/m), A is the area of the electrodes, and d is the distance between them. In this case, the area of the electrodes can be calculated using the formula A = πr^2, where r is the radius of the electrodes (which is half the diameter). So, the area of each electrode is A = π(2.0 cm)^2 = 4π cm^2. Substituting these values into the capacitance formula, we get C = (8.85 x 10^-12 F/m)(4π cm^2) / (0.0050 cm) = 2.22 x 10^-10 F.
To determine the charge on each electrode, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Plugging in the values we calculated above, we get Q = (2.22 x 10^-10 F)(200 V) = 4.44 x 10^-8 C.