Calculating Charge in a Parallel Connection of Capacitors

In summary, two capacitors with values of 11.1uF and 1.9uF are charged in parallel across a 333V battery. After being disconnected from the battery and each other, they are connected positive plate to negative plate and negative plate to positive plate. The resulting charge on the first capacitor is 0.004329C. When the capacitors are reconnected in series and then disconnected and connected positive plate to positive plate and negative plate to negative plate, the original charge on each capacitor can be calculated using the total charge and equivalent capacitance equations. The total charge on both capacitors when reconnected is the same as the original total charge, and the charge rearranges itself based on the same potential difference
  • #1
fenixbtc
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Homework Statement


Capacitors of 11.1uF and 1.9uF are charged as a parallel combination across a 333V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive to negative plate and negative to positive plate. Find the resulting charge on the first capacitor in units uC.


Homework Equations


1. Ceq = C1 + C2 parallel capacitance C1=11.1uF C2=1.9uF
2. Qtotal=V * Ceq

The Attempt at a Solution


1. Ceq = 11.1uF + 1.9uF = 13uF
2. Qtotal = 333 * 13uF = .004329 C

i tried using an equilibrium theorm:
(Qtot / Ceq) = (Qtot - q1) / 11.1uF -> .004329 / 13e-6 = (.004329 - q1) / 11.1e-6
with q1 = .0080253C but it's bigger than q= 333V * 11.1uF = .036963 so i don't think it's correct because that would be the max charge, right?

after this point do i treat it as capacitors in series or what?

part B of the problem states they are charged in series and then disconnected from the battery and connected positive plate to positive place and negative to negative plate

sorry if I'm overlooking something obvious.
thanks
david
 
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  • #2


Find the original charge on each capacitor. When they are reconnected, what's the total charge on both capacitors? When reconnected, what's the same across each capacitor? Use that to figure out how the charge rearranged itself upon reconnection.
 
  • #3


I would approach this problem by first understanding the concept of capacitance and how it relates to charge and voltage. Capacitance is the ability of a capacitor to store charge, and it is directly proportional to the amount of charge and inversely proportional to the voltage.

In this problem, we have two capacitors in parallel, which means they are connected to the same voltage source but have separate paths for current to flow. The total capacitance in a parallel connection is given by Ceq = C1 + C2, as you have correctly stated. This means that when the capacitors are disconnected from the battery and from each other, the total charge stored in both capacitors is the same as if they were connected in series.

Using this knowledge, we can find the total charge by using the equation Qtot = V * Ceq, where V is the voltage of the battery and Ceq is the equivalent capacitance of the parallel connection. Plugging in the values given in the problem, we get Qtot = 333 * 13uF = 4.329uC.

Now, to find the charge on the first capacitor, we can use the fact that the charge is divided between the two capacitors in proportion to their capacitance. This means that the first capacitor, with a capacitance of 11.1uF, will have a charge of q1 = (11.1/13) * 4.329uC = 3.663uC.

In part B of the problem, the capacitors are connected in series. This means that they are connected in a single path, and the same amount of current flows through both capacitors. In this case, the total capacitance in a series connection is given by 1/Ceq = 1/C1 + 1/C2. Using this equation, we can find the equivalent capacitance of the series connection to be Ceq = 6.9uF. Using the equation Qtot = V * Ceq, we get Qtot = 333 * 6.9uF = 2.301uC.

Since the capacitors are now connected in series, the charge on each capacitor will be the same. This means that the charge on the first capacitor will be q1 = 2.301uC / 2 = 1.1505uC.

In summary, the resulting charge on the first capacitor in part A is 3.663uC,
 

1. How do you calculate the total charge in a parallel connection of capacitors?

The total charge in a parallel connection of capacitors is equal to the sum of the individual charges on each capacitor. This means that you can simply add together the charges on each capacitor to get the total charge.

2. What is the formula for calculating the total charge in a parallel connection of capacitors?

The formula for calculating the total charge in a parallel connection of capacitors is Qtotal = Q1 + Q2 + Q3 + ... + Qn, where Qtotal is the total charge, Q1, Q2, Q3, etc. are the individual charges on each capacitor, and n is the number of capacitors in the parallel connection.

3. How does the total charge in a parallel connection of capacitors compare to the total charge in a single capacitor?

The total charge in a parallel connection of capacitors will always be greater than the total charge in a single capacitor. This is because the capacitors in a parallel connection are connected in such a way that the same voltage is applied across each capacitor, resulting in a greater charge being stored on each individual capacitor.

4. Can you use the same formula for calculating total charge in a parallel connection of capacitors if they have different capacitances?

Yes, the same formula can be used regardless of the capacitance values of the individual capacitors. As long as they are connected in parallel, the total charge will be the sum of the individual charges on each capacitor.

5. How does the total charge in a parallel connection of capacitors affect the overall voltage?

The total charge in a parallel connection of capacitors does not affect the overall voltage. The voltage across each capacitor will be the same as the voltage of the power source, regardless of the number or values of the capacitors. However, the total charge will increase with each additional capacitor in the parallel connection.

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