Let's break down the given problem systematically. We have three capacitors connected in series, and the capacitances are $C_1 = 8.5 \, \mu\text{F}$, $C_2 = 9.2 \, \mu\text{F}$, and $C_3$ is unknown. The charge on each capacitor is $Q = 11 \, \mu\text{C}$, and we are tasked with finding the potential at point 3, $V_3$.

Step-by-Step Solution

Step 1: Understanding the series connection of capacitors

For capacitors in series:

• The charge on each capacitor is the same, $Q = 11 \, \mu\text{C}$.
• The total voltage across the series arrangement is the sum of the voltages across each capacitor.
• The voltage across each capacitor $V = Q / C$, where $C$ is the capacitance of that particular capacitor.

Step 2: Calculating the voltage across each capacitor

We can start by calculating the voltage across the first capacitor $C_1$ using the formula $V = Q / C$: $V_1 = \frac{Q}{C_1} = \frac{11 \, \mu\text{C}}{8.5 \, \mu\text{F}} = 1.294 \, \text{V}$

Similarly, calculate the voltage across $C_2$ (the second capacitor): $V_2 = \frac{Q}{C_2} = \frac{11 \, \mu\text{C}}{9.2 \, \mu\text{F}} = 1.196 \, \text{V}$

Step 3: Voltage difference between the capacitors

The voltage across each capacitor is the potential difference. From the provided data, we have:

• $V_1 - V_2 = \Delta V_{12}$.
• $V_2 = 7.710 \, \text{V}$, which means the potential at point 2 is 7.710 V.

Since $V_2 = 7.710 \, \text{V}$ and $V_2 = V_1 - V_2$, let's calculate $V_3$ for the third capacitor.

Step 4: Calculate $V_3$

To compute the potential at point 3, we need to calculate how much potential change occurs due to the third capacitor (C3) using its charge formula. Keep in mind these data