Question

Two 3.0 cm × 3.0 cm square aluminum electrodes, spaced 0.50 mm apart, are connected to a 400 v battery.

(a) What is the capacitance?
(b) What is the charge on the positive electrode?

Answer 1

The capacitance of the system is 160 nF and the charge on the positive electrode is 64 µC.

Step-by-step explanation:

(a) To find the capacitance, we can use the formula C = εA/d, where ε is the permittivity of free space, A is the area of the electrodes, and d is the distance between them. In this case, the area of each electrode is (3.0 cm) x (3.0 cm) = 9.0 cm2 = 9.0 x 10-4 m2 and the distance between them is 0.50 mm = 0.50 x 10-3 m. Plugging these values into the formula, we get:

C = (8.85 x 10-12 F/m)(9.0 x 10-4 m2)/(0.50 x 10-3 m) = 0.16 x 10-6 F = 160 nF

(b) To find the charge on the positive electrode, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. In this case, the capacitance is 160 nF and the voltage is 400 V. Plugging these values into the formula, we get:

Q = (160 x 10-9 F)(400 V) = 64 x 10-6 C = 64 µC