To calculate the charge (\(Q\)) that flows from a battery to a capacitor, you can use the formula:
\[ Q = C \times V \]
where:
- \(Q\) is the charge in coulombs (C),
- \(C\) is the capacitance in farads (F),
- \(V\) is the voltage in volts (V).
Given:
- \(C = 9.00 , \mu\text{F} = 9.00 \times 10^{-6} , \text{F}\)
- \(V = 3 , \text{V}\)
Now, substitute the values into the formula:
\[ Q = (9.00 \times 10^{-6} , \text{F})(3 , \text{V}) = 27.0 \times 10^{-6} , \text{C} = 2.7 \times 10^{-5} , \text{C} \]
Therefore, the amount of charge that flows from the 3 V battery when connected to a 9.00 microfarad capacitor is
\[ \text{2.7} \times 10^{-5} , \text{C} \]
So the correct answer is:
2.7x10-5 C