(a) During a Physics Lab experiment, 1st year SFY students analyzed the behavior of capacitors by connecting two capacitors to power sources. In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF the capacitor to a 562 V source. Once these capacitors are charged, they are then removed from the power sources and are interconnected by connecting the two positive plates (of the capacitors) together, and the two negative plates together. Determine the potential difference
and charge across each capacitor.
(b) In the second part of the lab activities, students charged another capacitor (𝐶1) using a 165 V source. Once charged, the capacitor was then removed from the source and then connected to another capacitor (𝐶2) which was initially uncharged. If the students measured the final potential difference across each capacitor to be 15V, determine the value of (𝐶2).
(a) To determine the potential difference and charge across each capacitor, we can use the principle of conservation of charge. According to this principle, the total charge before and after the capacitors are interconnected remains the same.
Let's first calculate the charge on the 2.50 µF capacitor connected to the 746 V source.
Q1 = C1 * V1
Q1 = (2.50 µF) * (746 V)
Q1 = 1.865 mC
Similarly, let's calculate the charge on the 6.80 µF capacitor connected to the 562 V source.
Q2 = C2 * V2
Q2 = (6.80 µF) * (562 V)
Q2 = 3.8216 mC
Once the capacitors are interconnected, the charges add up, so the total charge is the sum of Q1 and Q2.
Qtot = Q1 + Q2
Qtot = 1.865 mC + 3.8216 mC
Qtot = 5.6866 mC
Since the capacitors are connected in parallel, the potential difference across each capacitor is the same and equal to the final potential difference.
Now, let's find the potential difference across each capacitor.
Vfinal = 746 V - 562 V
Vfinal = 184 V
Since the potential difference across each capacitor is the same, it is equal to 184 V.
To find the charge on each capacitor, we can use the formula Q = C * V.
For the 2.50 µF capacitor:
Q1 = (2.50 µF) * (184 V)
Q1 = 0.46 mC
For the 6.80 µF capacitor:
Q2 = (6.80 µF) * (184 V)
Q2 = 1.2512 mC
Therefore, the potential difference across each capacitor is 184 V, and the charge on the 2.50 µF capacitor is 0.46 mC, while the charge on the 6.80 µF capacitor is 1.2512 mC.
(b) In this part of the lab, we need to determine the value of C2.
When the capacitors are connected in series, the total charge remains the same, but the potential difference across each capacitor is different.
Let's assume the charge on both capacitors after they are interconnected is Q. The potential difference across C1 is 15 V, and the potential difference across C2 is also 15 V. Since the charge is the same, we can write the equation:
Q = C1 * V1 = C2 * V2
We already know the values of C1 (given as the initial charged capacitor) and V1 (measured potential difference across C1). Therefore, we can rearrange the equation to solve for C2:
C2 = (C1 * V1) / V2
Plugging in the values:
C2 = (C1 * V1) / V2
C2 = (165 V * 𝐶1) / 15 V
Now, we would need the value of C1 to proceed with the calculation.