Capacitors disconnected from the battery

In summary, two capacitors, C1 = 10 µF and C2 = 23 µF, are charged to 9V and then disconnected from the battery without changing the charge on the capacitor plates. When they are connected in parallel, with the positive plate of C1 connected to the negative plate of C2 and vice versa, the charges on each capacitor are not 9V. The charge on C1's positive plate is 207 µC and its negative plate is 0 µC. The charge on C2's positive plate is 0 µC and its negative plate is 90 µC. The net charge on the connected plates is 90 µC.
  • #1
laduch
3
0
Capacitors C1 = 10 µF and C2 = 23 µF are each charged to 9 V, then disconnected from the battery without changing the charge on the capacitor plates.
The two capacitors are then connected in parallel, with the positive plate of C1 connected to the negative plate of C2 and vice versa.
Afterward, what are the charge on and the potential difference across each capacitor?

The question ask for Q1, Q2(charge (µC)), and V1, V2(potential difference (V)).

I used equation Q=CV which I think is the right equation to use... but there are some conceptual problems like what happens when "C1 connected to the negative plate of C2". When battery is disconnected from the circuit, how is the distribution of the voltage related to capacitors?
 
Physics news on Phys.org
  • #2
Yes, Q=CV is the right equation. You can use it to figure out the charges on each capacitor when they are at 9V.

what happens when "C1 connected to the negative plate of C2"

Charge will flow, until the two capacitors are at the same voltage.

When battery is disconnected from the circuit, how is the distribution of the voltage related to capacitors?
At that point, each capacitor has a 9V potential difference.
 
  • #3
I applied the equation...
Q1 = C1*V1 = 23 µF * 9 V = 207 µC
Q2 = C2*V2 = 10 µF * 9 V = 90 µC
The results are incorrect.
potential difference on each Capacitor is not 9V.
Kind confusing... X(
 
  • #4
laduch said:
I applied the equation...
Q1 = C1*V1 = 23 µF * 9 V = 207 µC
Q2 = C2*V2 = 10 µF * 9 V = 90 µC

Okay, good.
So that means C1 has a charge of ______ on the positive plate, and ______ on the negative plate?
Similary, C2 has ______ on the + plate and _____ on the - plate?

Next, imagine what happens if the + plate of C1 is connected to the - plate of C2 ... i.e. what will be the net charge on those 2 plates?

The results are incorrect.
potential difference on each Capacitor is not 9V.
Kind confusing... X(

It is 9V before the capacitors are connected together. It is something different after they are connected.
 

Related to Capacitors disconnected from the battery

1. What happens when a capacitor is disconnected from the battery?

When a capacitor is disconnected from the battery, it retains the charge it had when it was connected. However, the voltage across the capacitor decreases over time due to the leakage of charge through its dielectric material.

2. Are there any safety concerns when disconnecting a capacitor from the battery?

Yes, there can be safety concerns when disconnecting a capacitor from the battery. If the capacitor is not properly discharged, it can still hold a dangerous amount of charge and can cause electric shock if touched.

3. How long does it take for a capacitor to fully discharge after being disconnected from the battery?

The time it takes for a capacitor to fully discharge after being disconnected from the battery depends on its capacitance and the resistance in the circuit. It can range from a few seconds to several minutes.

4. Can a disconnected capacitor be recharged and used again?

Yes, a disconnected capacitor can be recharged and used again. However, the quality of the capacitor may decrease over time due to the degradation of its dielectric material.

5. Is it necessary to discharge a capacitor before disconnecting it from the battery?

Yes, it is recommended to discharge a capacitor before disconnecting it from the battery to ensure safety and prevent damage to the capacitor. This can be done by shorting the terminals of the capacitor with a resistor or using a discharge tool.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
932
Replies
4
Views
401
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top