Answer:
[tex]4.1 \mu F[/tex]
Explanation:
In this problem, the capacitors C1 and C2 are connected in series to the battery.
The voltage of the battery, which is variable, is V.
The voltge across the capacitor C2 is V2: the graph of V2 versus V shows that a straight line with a slope of 0.45.
This means that the relationship between V and V2 is
[tex]V_2=0.45V[/tex] (1)
So, 45% of the voltage is on the capacitor C2; and therefore, 55% of the voltage must be on capacitor C1:
[tex]V_1=0.55V[/tex] (2)
When two capacitors are connected in series, they have the same charge on their plates, so
[tex]Q_1=Q_2[/tex]
Using the relationship between charge (Q), capacitance (C) and voltage (V) for a capacitor:
[tex]Q=CV[/tex]
We can rewrite the equation as
[tex]C_1V_1 =C_2V_2[/tex]
Or
[tex]\frac{C_1}{C_2}=\frac{V_2}{V_1}[/tex]
Substituting (1) and (2),
[tex]\frac{C_1}{C_2}=\frac{0.45V}{0.55V}[/tex]
And since
[tex]C_2=\mu F[/tex]
We find:
[tex]C_1=\frac{0.45}{0.55}C_2=0.818\cdot 5=4.1\mu F[/tex]
