Question

Two sheets of aluminum foil have the same area, a separation of 2.1 mm, and a capacitance of 1222 pF, and are charged to 12 V. Calculate the area of each sheet.2.90×10⁻¹ m². The separation is now decreased by 0.10 mm with the charge held constant. What is the new capacitance?

Answer 1

The new capacitance of the aluminum foil sheets when the separation decreases by 0.10 mm is approximately 1283 pF.

Step-by-step explanation:

To calculate the new capacitance of the aluminum foil sheets when the separation is decreased, we need to use the formula for the capacitance of a parallel plate capacitor C = (εA) / d, where C is the capacitance, ε is the permittivity of free space, A is the area of one plate, and d is the separation between the plates. The area of the sheets isn't needed for this part of the question, as it remains constant.

The initial capacitance was 1222 pF (1.222 x 10-12 F) when the separation was 2.1 mm (2.1 x 10-3 m), and the permittivity of free space is approximately 8.85 x 10-12 F/m. When the separation is decreased by 0.10 mm (0.10 x 10-3 m), the new separation becomes 2.0 mm (2.0 x 10-3 m). Substituting these values into the capacitor formula gives us:

C' = (εA) / d'

Since the charge is held constant and the capacitance is proportional to area and inversely proportional to separation, we can find the new capacitance by taking the ratio of the old separation to the new separation and multiply that by the old capacitance:

C' = C × (d / d') = 1.222 x 10-12 F × (2.1 x 10-3 m / 2.0 x 10-3 m) = 1.2831 x 10-12 F

The new capacitance is therefore approximately 1283 pF.