Parallel Capacitors: Find Energy Decrease

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Homework Help Overview

The problem involves two capacitors connected in parallel to a battery, with specific capacitance values and a voltage. After being disconnected from the battery, the capacitors are reconnected with opposite plates together, raising questions about the energy change in the system.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of charge on each capacitor and the implications of connecting opposite plates together. Questions arise about the net charge and the nature of the circuit after reconnection.

Discussion Status

Participants have made initial calculations regarding charge but express confusion about the reasoning behind the net charge and the configuration of the circuit. Some guidance has been provided regarding the addition of charges and the use of energy formulas, but no consensus has been reached on the overall understanding.

Contextual Notes

There is mention of a mix-up between capacitance and charge units, indicating potential misunderstandings about the relationships between the variables involved.

wcelectric
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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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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?
 
So would I just add them?

I am confused about the plates of opposite signs being together. It's still a parallel circuit?
 
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.
 
Ok, so the net charge is 140 μC. But I don't understand why.
 
Last edited:
I'd use the formula
[tex]U = \frac{1}{2}\frac{Q^2}{C}[/tex]
 
Oops. I mixed up C with Coulombs.
 
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.
 

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