Potential Difference of C2: 9V Capacitors in Series

AI Thread Summary
To find the potential difference across capacitor C2 in a series circuit with capacitors C1 (6 μF) and C2 (12 μF) connected to a 9V battery, first calculate the equivalent capacitance of the series combination. The formula for equivalent capacitance in series is 1/C_total = 1/C1 + 1/C2. Once the total capacitance is determined, use the relationship Q = C_total * V_total to find the total charge in the circuit. Since the charge on capacitors in series is the same, the potential difference across C2 can be calculated using V = Q/C2. The solution requires determining both the equivalent capacitance and the total charge before finding the voltage across C2.
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Homework Statement


Two capacitors C1= 6 μF and C2 = 12 μF are connected in series, and the resulting combination is connected to a 9.00 V battery. What is the potential difference across C2?


Homework Equations



V=Q/C

The Attempt at a Solution



Q isn't given so I'm not sure what I'm suppose to do here.
 
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Since they are placed in series, what the equivalent capacitance of the system? Hence what is the total charge in the circuit?
 
the Q's are all equal...would it be zero?
 
tag16 said:
the Q's are all equal...would it be zero?

Right so since, the charge on both capacitors is Q. What is the total capacitance of the series circuit C? When you get this value, what is the value of the total charge Q?
 

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