Question

The charge on a 3.7μF capacitor decreases by 66μC. If the initial voltage across the capacitor is 62 V, determine the final voltage. V

f = Part B Two parallel plates each have an area of 26.3 cm

Answer 1

The final voltage across the capacitor after the charge decreases by 66μC is approximately 44.16V, calculated by the relationship between charge, capacitance, and voltage.

Step-by-step explanation:

To determine the final voltage across the capacitor after the charge decreases by 66μC, we can use the relationship between charge (Q), capacitance (C), and voltage (V): Q = CV. So the change in voltage (ΔV) can be calculated using the decrease in charge (ΔQ) divided by the capacitance (C).

Initial Voltage: 62V

Capacitance (C): 3.7μF = 3.7 x 10-6F

Decrease in Charge (ΔQ): 66μC = 66 x 10-6C

Step 1: Calculate the change in voltage (ΔV) using the formula ΔQ = C x ΔV: ΔV = ΔQ / C ΔV = 66 x 10-6C / 3.7 x 10-6F ΔV ≈ 17.84V

Step 2: To find the final voltage (Vf), subtract the change in voltage (ΔV) from the initial voltage (Vi): Vf = Vi - ΔV Vf = 62V - 17.84V Vf ≈ 44.16V

Therefore, the final voltage across the capacitor is approximately 44.16V after the charge decreases by 66μC.