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Capacitor 3 in Figure (a) is a variable capacitor; its capacitance $C_3$ can be varied. Figure (b) gives the electric potential $V_1$ across $C_1$ versus $C_3$. The horizontal scale is set by $C_3 = 20 \, \text{μF}$. Electric potential $V_1$ approaches an asymptote of $10 \, \text{V}$ as $C_3$ approaches infinity.

What are

(a) the electric potential $V$ across the battery,

(b) $C_1$, and

(c) $C_2$?

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equations: $C=\frac{q}{V}=\frac{A}{\epsilon_0 d}$

How does one solve for the voltage on the battery, which is supposed to be $10 \, \text{V}$?

What I've tried to do is look at when $C_3 = \infty$ which means that the distance between the capacitor is 0, so its just a wire. Then I get a parallel circuit with $C_1$, and $C_2$, where the voltage going through $C_1$ is $10 \, \text{V}$. This makes a series circuit with $C_1$, $C_2$. So I know that the voltage of the battery is $10 \, \text{V}$ + the voltage across $C_2$. I can't figure out how to get $V$ for the battery.

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When $C3=\infty$, when you combine it in parallel with $C2$ you have infinite capacitance, which means zero voltage. When you combine that in series with $C1$ you just have $C1$, so all the battery voltage is across $C1$

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