Three identical capacitors are connected in series to a battery. If a total charge of Q flows from the battery, how much charge does each capacitor carry?
A)9Q
B)Q/3
C)Q
D)3Q
E)Q/9
The +Q charge from the + side of the battery actually goes to the + side of the first capacitor. That is really the end of the story so the answer is C)
-------
but I will babble on.
an equal and opposite - Q charge appears on the - side of that first capacitor. It comes from the top side of capacitor 2 leaving + Q on the + side of capacitor 2
that draws an -Q from the top of capacitor 3, leaving on the top and -Q on the bottom of capacitor 3
--------
the end
When capacitors are connected in series, the charge on each capacitor is the same. Therefore, if a total charge of Q flows from the battery, each capacitor carries Q/3 charge.
The correct answer is B) Q/3.
To find out how much charge each capacitor carries in a series circuit, we need to use the concept of equivalent capacitance. In a series circuit, the capacitors are connected end-to-end, so the charge flowing through each capacitor is the same.
The equivalent capacitance (C_eq) of capacitors connected in series can be calculated using the formula:
1/C_eq = 1/C1 + 1/C2 + 1/C3
Since the three capacitors are identical, we can simplify the equation as follows:
1/C_eq = 1/C + 1/C + 1/C = 3/C
Multiply both sides by C to get:
1/C_eq = 3/C
Now, let's find the value of C_eq. Since we have three identical capacitors, we can represent the capacitance of each capacitor as C.
1/C_eq = 3/C
Multiply both sides by C_eq to get:
C_eq = C/3
So, the equivalent capacitance of the series circuit is C/3.
Now, we know that the total charge flowing from the battery is Q. In a series circuit, the total charge flowing through the circuit is equal to the charge on each capacitor.
Therefore, the charge carried by each capacitor is Q.
Hence, the answer is C) Q.