Calculating Capacitor Charge and Potential Difference in Series

In summary, when a charged capacitor C1 is connected directly to an uncharged capacitor C2, the charge on each capacitor will be Q1 = Q0C1/(C1 + C2) and Q2 = Q0C2/(C1 + C2). The potential difference across each capacitor will be V = Q0 / (C1 + C2). This is determined by the fact that the voltage around the loop is 0 in equilibrium, and the total charge remains constant. The charges on the capacitors can be expressed as Q_1 = C_1V and Q_2 = C_2V, and the total charge is Q_0 = Q_1 + Q_2. Therefore, by
  • #1
Soaring Crane
469
0
A capacitor C1 carries a charge Q0. Is is then connected directly to a second, uncharged, capacitor C2.

What charge will each carry now? Q1 = Q0C1/(C1 + C2); Q2 = Q0C2/(C1 + C2)


What will be the potential difference across each?
V = Q0 / (C1 + C2)

-------||-------||--------
|*****C1*****C2******|
|********************|
|********************|
|********************|
|_______________________|

I know that the formula Q = CV is used, but how do you manipulate it (other than V = Q/C) to get the answers in bold?

Thanks.
 
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  • #2
The voltage around the loop will be 0 in equilibrium, and whatever charge the second capacitor gets, the first one must lose. Does that help at all?
 
  • #3
Soaring Crane said:
I know that the formula Q = CV is used, but how do you manipulate it (other than V = Q/C) to get the answers in bold?
Since the capacitors are connected in parallel, they have the same voltage (V). So the charges on the capacitors are: [itex]Q_1 = C_1V[/itex] and [itex]Q_2 = C_2V[/itex]. The total charge remains the same, so [itex]Q_0 = Q_1 + Q_2[/itex]. Solve for V.
 
  • #4
I think I understand what you're saying, but I don't know how to apply it to the questions.
 
  • #5
Oh, OK. Let me try it.
 

1. What is a capacitor?

A capacitor is an electronic component that is used to store and release electrical charge. It is made up of two conductive plates separated by an insulating material, known as the dielectric.

2. How do I calculate the capacitance of a capacitor?

The capacitance of a capacitor can be calculated by dividing the charge stored on the plates by the voltage applied to the capacitor. It can also be calculated by multiplying the permittivity of the dielectric material by the area of the plates and dividing by the distance between them.

3. What is the role of a capacitor in a circuit?

A capacitor has several important roles in a circuit. It can act as a filter, smoothing out fluctuations in voltage. It can also be used to store and release energy, and in some cases, it can even act as a temporary power source.

4. How does the capacitance of a capacitor affect its performance?

The capacitance of a capacitor directly affects its ability to store charge and release energy. A larger capacitance allows for more charge to be stored, while a smaller capacitance may result in a faster discharge time. The type of dielectric material used also impacts the performance of a capacitor.

5. What are some common applications of capacitors?

Capacitors have a wide range of applications in various electronic devices and systems. They are commonly used in power supplies, audio equipment, and electronic filters. They can also be found in computer circuits, electric motors, and many other electronic devices.

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