Solve "Charging a Capacitor" Homework Problem

• clickcaptain
In summary, the problem involves a 4.6 mu or micro FF capacitor initially charged to 15.0 V and then connected in series with an uncharged capacitor. The potential difference across the first capacitor drops to 7 V, and the question is to find the capacitance of the second capacitor. Through calculations of charge and voltage, the solution is found to be 9.9 microF, with the capacitors being connected in parallel.
clickcaptain

Homework Statement

A 4.6 mu or micro FF capacitor is charged to a potential difference of 15.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 4.6 mu or micro FF capacitor then drops to 7 V. What is the capacitance of the second capacitor?

Homework Equations

Q = CV

Capacitor in series
1/Ceq= 1/C1+1/C2

The Attempt at a Solution

I found the charge of the capacitor in the first scenario when the potential difference is 15 V

Q = CV
= 4.6E-6 F *15 V
= 7.35E-5 C

Since the potential drops to 7 V, you know the capacitors are in a series so that means the charge remains the same on both capacitors but potential changes

Capacitor in series
1/Ceq= 1/C1+1/C2

Q = Ceq * V

7.35E-5 C = (1/(4.6E-6 F) +1/C2)*7 V

I solved for C2...

C2 = -4.6E-6 C

- which is negative so its obvious that its wrong, but could someone tell me where I went wrong that would be greatly appreciated. Thank you.

I think there is an error in your calculation of the charge on the first capacitor. It seems that your work is correct, but when I multiply (4.6 microF)*(15V) I get the charge to be
6.9e-5 C. Since you know that the voltage drops to 7V across the first capacitor, you can treat the circuit as if there is initially a 7V battery connected to only the second capacitor, which is unknown. In this case, you know that V=7V and Q= 6.9e-5C. From here you can manipulate the equation so that C=(Q/V). Therefore, C should equal 9.9 microF.

Your way makes sense, and its a lot simpler than what I tried. The answer didn't work though so I'm going to try to keep double checking figures, i can't figure out what else would be wrong.

Aren't the capacitors connected in parallel? And isn't the amount of charge calculated in the first part the total amount of charge in the 2-capacitor system in part two?

clickcaptain,
I'm pretty sure you were doing it correctly in your first post (except for the first charge calculation, I got 6.9*10-5 C like w3390). Draw out the 2-capacitor circuit in the way it's described and I think you'll see that they are in parallel. When calculating C2, your Ceq should be Ceq = (C1 + C2) and not 1/Ceq = (1/C1) + (1/C2).

Are you arriving at the correct solution now?

1. What is a capacitor?

A capacitor is a device that stores electrical energy by accumulating an imbalance of electric charge between two conductive plates separated by an insulating material.

2. How does charging a capacitor work?

Charging a capacitor involves connecting it to a power source, such as a battery, causing electrons to flow from one plate to the other, building up a charge on the capacitor. The amount of charge is determined by the capacitance of the capacitor and the voltage of the power source.

3. What is the equation for calculating the charge on a capacitor?

The equation for calculating the charge on a capacitor is Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor in volts.

4. How do you solve a charging capacitor homework problem?

To solve a charging capacitor homework problem, you will need to determine the capacitance of the capacitor, the voltage of the power source, and the amount of time that has passed since the capacitor was connected to the power source. You can then use the equation Q = CV to calculate the charge on the capacitor at that specific time.

5. What are some common mistakes when solving a charging capacitor problem?

Some common mistakes when solving a charging capacitor problem include forgetting to convert units to the correct form (e.g. converting microfarads to farads), not taking into account the initial charge on the capacitor, and not considering the effects of resistance in the circuit. It is important to carefully check all calculations and make sure all necessary variables are included in the equation.

• Introductory Physics Homework Help
Replies
7
Views
995
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
390
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
2K