Question

Part 10 of 10 - Going Further: Switching Polarity, Part 1 The partners now consider the same three capacitors in parallel, C1 = 3.25 uF, C2 = 3.95 uF, and C3 = 8.85 uF initially connected to a potential difference of ∆Vb = 35.0 V. Kylie suggests removing the battery and removing and replacing C2, this time with positive and negative sides switched. The capacitors form an isolated system with the same capacitance as before. She asks Blake how he can determine the new potential difference across the capacitor combination. What is his response? "Because C2 is reversed, the charge on it is zero, and because Q = C∆V, Qnew = (C1 + C3)∆V. Solve this for ∆V." "Because C2 is reversed, the charge on it is twice the original charge, so Qnew = (C1 + 2C2 + C3)∆V. Solve this for ∆V." "Because the system is isolated, Qnew = Q1 - Q2 + Q3, and because Q = C∆V, Qnew = (C1 + C2 + C3)∆V. Solve this for ∆V." "Because the system is isolated, Qnew = Q1 - Q2 - Q3, and because Q = C∆V, Qnew = (C1 + C2 + C3)∆V. Solve this for ∆V." Blake is correct. The charge will rearrange itself so that the potential difference across each capacitor is equal. Going Further: Switching Polarity, Part 2 Blake challenges Kylie to calculate the new voltage. What is the new voltage?

Answer 1

When capacitors are connected in parallel, the total capacitance is the sum of individual capacitances. In this scenario, switching the polarity of one capacitor results in a new potential difference determined by the sum of the charges on the remaining capacitors. The equation Q = C∆V can be used to calculate the new potential difference.

Step-by-step explanation:

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. In this case, C1 = 3.25 uF, C2 = 3.95 uF, and C3 = 8.85 uF. So the total capacitance is Cp = C1 + C2 + C3. However, when C2 is reversed, the charge on it becomes zero. Therefore, the new total charge is Qnew = Q1 + Q3, where Q1 and Q3 are the charges on C1 and C3, respectively.

The relationship between charge, capacitance, and potential difference is given by Q = C∆V. So, Qnew = (C1 + C3)∆V. To find the new potential difference, we can rearrange the equation to solve for ∆V: ∆V = Qnew / (C1 + C3).