Parallel Capacitors: Find Energy Decrease

In summary, the problem involves two capacitors connected in parallel with a battery of 28V. The capacitors have values of 9.0 μF and 4.0 μF. After being disconnected and reconnected with plates of opposite signs together, the net charge of the system is 140 μC. To find the energy decrease, the formula U = \frac{1}{2}\frac{Q^2}{C} can be used, with Q being the net charge and C being the equivalent capacitance of the parallel combination.
  • #1
wcelectric
5
0

Homework Statement



Two capacitors are connected in parallel to a battery with voltage of 28V. Let [tex]C_1 = 9.0 \mu F[/tex] and [tex]C_2 = 4.0 \mu F [/tex].
Suppose the charged capacitors are disconnected from the source and from each other, and then reconnected to each other with plates of opposite signs together. By how much does the energy of the system decrease?


Homework Equations


[tex]u=\frac{1}{2}CV^2[/tex]

[tex]C=\frac{Q}{V}[/tex]


The Attempt at a Solution



I have [tex]Q_1 = 252 \mu C[/tex] and [tex]Q_2 = 112 \mu C[/tex]. Beyond that I'm stuck.
 
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  • #2
wcelectric said:

The Attempt at a Solution



I have [tex]Q_1 = 252 \mu C[/tex] and [tex]Q_2 = 112 \mu C[/tex]. Beyond that I'm stuck.

That's a reasonable start.

Next, they say that the plate with +252 μC plate is connected to the -112 μC plate on the other capacitor. How much net charge would there be, being shared between these two plates?
 
  • #3
So would I just add them?

I am confused about the plates of opposite signs being together. It's still a parallel circuit?
 
  • #4
wcelectric said:
So would I just add them?
Yes.
I am confused about the plates of opposite signs being together. It's still a parallel circuit?
Yes.
 
  • #5
Ok, so the net charge is 140 μC. But I don't understand why.
 
Last edited:
  • #6
I'd use the formula
[tex]U = \frac{1}{2}\frac{Q^2}{C}[/tex]
 
  • #7
Oops. I mixed up C with Coulombs.
 
  • #8
wcelectric said:
Ok, so the net charge is 140 μC. But I don't understand why.

It might help to think of an atom, say hydrogen for example. It contains two charged particles:

(1) a proton with charge 1.6 × 10-19 C
(2) an electron with an opposite charge of -1.6 × 10-19 C

To get the net charge of the hydrogen atom, we just add up all the charges in the atom, which is zero net charge in this example.

Similarly, whenever you combine charges, you simply add up the separate charges to get a net total charge.

diazona said:
I'd use the formula
[tex]U = \frac{1}{2}\frac{Q^2}{C}[/tex]

That's right. You have Q, and just need to figure out what C is for this parallel capacitor combination.
 

1. How do parallel capacitors affect energy decrease?

In parallel capacitors, the total energy stored is the sum of the energies stored in each capacitor. Therefore, the energy decrease is also affected by the individual capacitors' values and how they are connected.

2. How do I calculate the total energy decrease in parallel capacitors?

To calculate the total energy decrease in parallel capacitors, you can use the formula E = 1/2 * C * V^2, where E is the energy, C is the total capacitance, and V is the voltage across the capacitors.

3. Can I use parallel capacitors to increase energy decrease?

No, parallel capacitors do not increase energy decrease. The total energy stored in parallel capacitors is the same as the individual capacitors, but the energy decrease may be distributed differently.

4. How do the individual capacitors' values affect energy decrease in parallel capacitors?

The individual capacitors' values determine the total capacitance in parallel capacitors, which directly affects the total energy stored and the energy decrease. A higher capacitance value will result in a higher energy decrease.

5. Are there any limitations to using parallel capacitors for energy decrease?

Yes, there are limitations to using parallel capacitors for energy decrease. One limitation is that the capacitors must have the same voltage rating, otherwise, they may not operate properly and could potentially cause damage. Additionally, the total capacitance in parallel capacitors cannot exceed the maximum capacitance rating of the circuit.

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