In this setup, we can use the relationship between the total voltage $V$, the voltage across $C_2$ (which is $V_2$), and the capacitances $C_1$ and $C_2$ when they are connected in series.

Step 1: Voltage division in series capacitors

For capacitors $C_1$ and $C_2$ connected in series, the voltage across each capacitor is related to the total voltage $V$ by the following voltage division rule:

$\frac{V_2}{V} = \frac{C_1}{C_1 + C_2}$

Here, $V_2$ is the voltage across $C_2$, and $V$ is the total voltage applied across the series combination.

Step 2: Linear relationship between $V_2$ and $V$

From the given information, the plot of $V_2$ versus $V$ results in a straight line, implying that:

$V_2 = m \cdot V$

Where $m$ is the slope of the line, which is given as $0.650$. This slope represents the fraction of the total voltage that appears across $C_2$:

$m = \frac{C_1}{C_1 + C_2}$

Step 3: Solve for $C_1$

Now, we can solve for $C_1$ using the known values for $m$ and $C_2$:

$0.650 = \frac{C_1}{C_1 + 5.00\, \mu F}$

Multiply both sides by $C_1 + 5.00\, \mu F$ to get:

$0.650 \cdot (C_1 + 5.00\, \mu F) = C_1$

Expanding:

$0.650 \cdot C_1 + 3.25\, \mu F = C_1$

Now, isolate $C_1$ on one side:

$3.25\, \mu F = C_1 - 0.650 \cdot C_1$

$3.25\, \mu F = 0.350 \cdot C_1$

Solve for $C_1$:

$C_1 = \frac{3.25\, \mu F}{0.350}$

$C_1 = 9.29\, \mu F$

Final Answer:

The capacitance $C_1$ is approximately 9.29 μF.

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Would you like more details on this solution or have any other questions?

Here are some related questions that you might find interesting:

1. How does the series combination of capacitors affect the overall capacitance?
2. What would happen if the capacitors were connected in parallel instead of series?
3. How can you calculate the total capacitance when capacitors are connected in series?
4. What is the relationship between the voltage across capacitors in parallel and series circuits?
5. Why does the slope of the $V_2$ versus $V$ graph give us information about the capacitance?

Tip: When dealing with series capacitors, remember that the total capacitance is always smaller than the smallest individual capacitor in the series.