Answer :
Final answer:
Unfortunately, we cannot determine the exact amount of charge stored on the plates without knowing the capacitance values of the individual capacitors in the combination.
Explanation:
To find the total charge stored on the plates of the capacitors, we need to know the capacitance values of the individual capacitors in the combination. The given question does not provide this information, so we cannot calculate the exact charge.
However, if we are given the capacitance values, we can use the formula Q = CV to calculate the charge. We would multiply the capacitance of each capacitor by the voltage across the combination (9.00 V) to find the charge stored on each capacitor. Then, we would sum up the charges to find the total charge stored on the plates.
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The charge stored on the plates of the capacitors is [tex]\( 297 \times 10^{-6} \) C[/tex] which can also be expressed as [tex]\( 2.97 \times 10^{-4} \) C[/tex]
Use the formula [tex]\( Q = CV \)[/tex], where [tex]\( Q \)[/tex] is the charge, [tex]\( C \)[/tex] is the capacitance, and [tex]\( V \)[/tex] is the voltage across the capacitors.
The capacitors are connected in parallel, so the total capacitance [tex]\( C_{\text{eq}} \)[/tex] is the sum of the individual capacitances.
Given:
[tex]- \( C_1 = 15 \, \mu F = 15 \times 10^{-6} \, F \)[/tex]
[tex]- \( C_2 = 12 \, \mu F = 12 \times 10^{-6} \, F \)[/tex]
[tex]- \( C_3 = 6 \, \mu F = 6 \times 10^{-6} \, F \)[/tex]
The equivalent capacitance [tex]\( C_{\text{eq}} \)[/tex] is:
[tex]\[ C_{\text{eq}} = C_1 + C_2 + C_3 \][/tex]
[tex]\[ C_{\text{eq}} = 15 \times 10^{-6} + 12 \times 10^{-6} + 6 \times 10^{-6} \][/tex]
[tex]\[ C_{\text{eq}} = 33 \times 10^{-6} \, F \][/tex]
Now, we can calculate the charge [tex]\( Q \)[/tex] stored on the plates using [tex]\( Q = C_{\text{eq}} \times V \), where \( V = 9.00 \, V \)[/tex]
[tex]\[ Q = (33 \times 10^{-6}) \times 9.00 \][/tex]
[tex]\[ Q = 297 \times 10^{-6} \, C \][/tex]
[tex]Q=\( 2.97 \times 10^{-4} \) C[/tex]
