Solving Capacitor Problem with 2 Capacitors

In summary: Because Q2 is 0, I crossed it out, but then I thought I should have left it there and put a 0, but then I thought that might have been confusing so I just deleted it. I should probably have left it!!!In summary, the problem involves finding the value of C2 when a 7.7 uF capacitor charged by a 125-V battery is connected to a second initially uncharged capacitor, resulting in a final voltage of 15 V on both capacitors. By applying the law of conservation of electric charge, it is found that C2 is equal to approximately 6.4E-5 F.
  • #1
thisisfudd
56
0
Hi,

I understand generally how to do problems with capacitors, but I am faced here with a problem about 2 capacitors and I am not sure what the appropriate calculations are for combining their capacitance, etc. Here is the problem:

A 7.7 uF capacitor is charged by a 125-V battery and then is disconnected from the battery. When this capacitor is the connected to a second (initially uncharged) capacitor, C2, the final voltage on each capacitor is 15 V. What is the value of C2?

So here is what I did:

Q = C V
Q = 7.7E-6 F x 125 V
Q = 9.625E-4 C per plate (first capacitor)

PE = 1/2 Q^2/C
PE = 1/2 x (9.625E-4)^2/7.7E-6 F
PE = .06

From this, I used PE = 1/2 CV^2
.06 = 1/2 (7.7E-6 F + x)(15^2)
To get: 5.3E-4 F which I am pretty sure is way too large a number
 
Physics news on Phys.org
  • #2
Are u sure it's not [itex] C_{2}\sim 5.3\cdot 10^{-5}F [/itex] ...?

I got that by applying the law of conservation of electric charge (the battery is disconnected when the second capacitor is brought in)...

Daniel.
 
  • #3
Hi,

How did you get the answer you are giving? I am not sure I understand. If the battery is disconnected, C1 retains the charge it has from the battery, or 9.625E-4 C (per plate, right?). From what you said, I would do the following:

The charge is going to be the same now because there is no other battery. So then that would be:

Q = CV

Would you multiply times 2 because there are two plates?

.001925 = C x (15 + 15)

So then C = 6.42E-5

Is that right or totally wrong?
 
  • #4
Nope.The charge would move from one capacitor to another,as to insure equal potentials...

Daniel.
 
  • #5
Hi,
OK, I'm not sure I understand. So they would have equal potentials. So the total charge would be 1.625E-4. The voltage on each battery would be 15 (total = 30). So then I just need to find C2.

Ctotal = C1 + C2
C1 =7.7 uF
So C2 = Ctotal - C1

Is that it?
 
  • #6
Q=Q1+Q2 Charge conservation.What is C total...?

Daniel.
 
  • #7
OK, concentration of electric charge. So you have the electric charge from the first part, and then you have that total charge for the second part, right? Because there is no other source, because there is no battery.

I was thinking that each capacitor has its own capacitance, for C1 it is 7.7 uF and for C2, well that is what we are trying to find.

What is Q = Q1 + Q2? So each capacitor gets half of the charge, so each capacitor gets 9.625E-4/2. So do I do:

Q = CV
(9.625E-4/2) = C2 x 15V

??
 
  • #8
No.Thay would get help of the charge,iff the capacitors would have the same C,but my guess it they don't.

Daniel.

P.S.Provide a number for C_{2}
 
  • #9
How can I provide a number for C2 if that is what we are looking for?

So if they don't each get half the charge (I thought they did?) then don't we have two variables?

Because then we have Q = Q1 + Q2 AND C = C1 + C2 (I am assuming that there are two capacitances, yes?). So we know V = 30.

We know that C2 is initially 0. And we know that C1 is initially 7.7 pF. And we know that the final voltage. And we know the final TOTAL charge but not the charge for each individual capacitor.

I also don't understand what the voltage thing means. If the voltage is the same on each, does that mean they have the same capacitance?

Sorry I'm starting to guess a little at this point. I am trying to think about it but I don't really understand.
 
  • #10
Q1=C1U
Q2=C2U

Add,factor U and extract C2.

Daniel.
 
  • #11
Thanks Daniel. Sorry I'm a pain.
 
  • #12
There's not a problem,whatsoever;as a side thing,i gave you the hint in post #2...:tongue2:

Daniel.
 
  • #13
Now that I get what you were saying,

Q1 + Q2 = V (C1 + C2)
9.625E-4 + 0 = 15 (7.7E-9 F + x)
x= 6.4E-5?
 
  • #14
No.Check the numbers again.And the power is [itex] 10^{-6} [/itex]...:wink:

Daniel.
 
  • #15
Right, I was thinking pF ... this is just getting worse and worse. Anyhoo, let's try this again:

I know 9.625E-4 is right. And 15 is right, right? Because each of them gets multiplied by 15.

So 9.625E-4 + 0 = 15 (7.7E-6 + C2)

5.6E-5?

Please let that be right argh :cry:
 
  • #16
It looks okay.

Daniel.

P.S.What's with that "+0"...?
 
  • #17
Well the 0 was supposed to be for Q2, since Q2 is 0 (that is, the charge brought to the system by the second capacitor, that is how I am thinking about it even though that's not very physics-y, sounds more like a tupperware party).
 

1. What is a capacitor?

A capacitor is an electrical component that stores electric charge. It is composed of two conductive plates separated by an insulating material known as a dielectric.

2. How do I solve a capacitor problem with two capacitors?

To solve a capacitor problem with two capacitors, you first need to determine the equivalent capacitance. This can be done by using the formula Ceq = C1 + C2, where C1 and C2 are the individual capacitances of the two capacitors. Once you have the equivalent capacitance, you can use the formula Q = CV to calculate the charge on each capacitor and the total energy stored by both capacitors.

3. What is the difference between series and parallel capacitors?

In series capacitors, the capacitors are connected end-to-end, while in parallel capacitors, the capacitors are connected side-by-side. In series capacitors, the equivalent capacitance is calculated by using the formula 1/Ceq = 1/C1 + 1/C2, while in parallel capacitors, the equivalent capacitance is simply the sum of the individual capacitances.

4. How do I know which formula to use when solving a capacitor problem?

The formula you use will depend on the circuit configuration and the information given in the problem. If the capacitors are in series, use the series formula. If they are in parallel, use the parallel formula. If you are not sure, you can always use the general formula Q = CV to calculate the charge on each capacitor.

5. What are some real-world applications of capacitors?

Capacitors have a wide range of applications in electronics, including storing energy in flash cameras, smoothing power supply voltage in electronic devices, and filtering out noise in audio circuits. They are also used in electric motors, air conditioners, and power factor correction in industrial equipment.

Similar threads

Ranking charge stored on three capacitors by a battery
    • Introductory Physics Homework Help
Replies
6
Views
1K
Capacitor Questions Charging and Discharging
    • Introductory Physics Homework Help
Replies
4
Views
2K
Unknown circuit in a black box
    • Introductory Physics Homework Help
Replies
3
Views
717
Redistribution of Pre-Charged Capacitors
    • Introductory Physics Homework Help
Replies
11
Views
2K
Uncharged capacitors connected in series
    • Introductory Physics Homework Help
Replies
18
Views
1K
Potential difference between 2 points in a capacitor circuit
    • Introductory Physics Homework Help
Replies
30
Views
1K
Capacitors: find he charge for the new battery
    • Introductory Physics Homework Help
Replies
6
Views
994
Changes in capacitor after dielectric inserted
    • Introductory Physics Homework Help
Replies
17
Views
2K
Find the capacitance of a half-full capacitor
    • Introductory Physics Homework Help
Replies
12
Views
2K
Capacitors charged in parallel
    • Introductory Physics Homework Help
Replies
3
Views
1K

Hot Threads

Recent Insights

Back
Top