Redistribution of charges in capacitor

[AdityaDev]

Homework Statement

A capacitor of capacity C is charged with potential difference V. Another capacitor of capacity 2C is charged to 4V.they are connected with reverse polarity after removing batteries. The heat produced during redistribution of charges is...

Homework Equations

##Q = CV##
##U = 0.5CV^2##
[/B]

The Attempt at a Solution

Let Q be charge in capacitor with capacitance C and other be q initially.
##Q=CV##
##q=8CV##
##U1 = Q^2/2C + q^2/(4C)##
##U2 = (Q+q)^2/2(C+4C)##
U2 is final energy... Net charge is Q+q and since potential difference becomes same when connected... C(net) = C1+C2
Heat = delta U
##delta U = C1C2(V1-V2)/(C1+C2)##
Here V gets added up...( I don't know why... But I got the answer)
Hence ##U = C1C2(V1+V2)/(C1+C2)##

You will get correct answer but why V1 + V2 is there some other method?

 

[gneill]

Homework Statement

A capacitor of capacity C is charged with potential difference V. Another capacitor of capacity 2C is charged to 4V.they are connected with reverse polarity after removing batteries. The heat produced during redistribution of charges is...

Homework Equations

##Q = CV##
##U = 0.5CV^2##

3. The Attempt at a Solution [/B]
Let Q be charge in capacitor with capacitance C and other be q initially.
##Q=CV##
##q=8CV##
##U1 = Q^2/2C + q^2/(4C)##
##U2 = (Q+q)^2/2(C+4C)## ← One capacitor is C, the second is 2C
U2 is final energy... Net charge is Q+q and since potential difference becomes same when connected...

No, the total charge is not Q + q. The capacitors were connected in reverse polarity so there will be some charge cancellation.

C(net) = C1+C2
Heat = delta U
##delta U = C1C2(V1-V2)/(C1+C2)##
Here V gets added up...( I don't know why... But I got the answer)
Hence ##U = C1C2(V1+V2)/(C1+C2)##

You will get correct answer but why V1 + V2 is there some other method?

Heat is dissipated by current flowing through resistance. If you place an arbitrary resistor in the circuit to represent the small resistance in the wiring and integrate the power, ##I^2R## for t = 0 to ∞, then you have another way to find the result. Note that the R will conveniently disappear in the working out (try it, you'll see). Of course this assumes that you know what the current will look like for an RC circuit...

 

[AdityaDev]

No, the total charge is not Q + q. The capacitors were connected in reverse polarity so there will be some charge cancellation.

From conservation of charges net charge will always be Q + q. But if q is negative then there can be cancelation. I'll take sign later.
If +CV is connected to -8CV then some charge will flow from cv to 8cv.
Similarly a charge will flow from +8CV to -CV... This will happen till potential across capacitors become same.

 

[gneill]

From conservation of charges net charge will always be Q + q. But if q is negative then there can be cancelation. I'll take sign later.

When you charge a capacitor with a charge Q, then there will be +Q on one plate and -Q on the other plate. Net charge on the entire device is zero. When you connect two capacitors, reversing the polarity of one of them, there's an opportunity for the + and - charges that meet each other through the connecting wire to cancel and "vanish" from the system. The overall net charge in the system is still zero, so the conservation gods are appeased. But the separate charges remaining to distribute on the joined plates has decreased.

 

[HalfLostAndSearching]

I would like to provide a more comprehensive explanation for the redistribution of charges in a capacitor. A capacitor is a passive electronic component that stores electrical energy in the form of an electric field. When a capacitor is charged, it accumulates equal and opposite charges on its plates, creating a potential difference between them.

In the given scenario, we have two capacitors of different capacitances (C and 2C) and different initial potential differences (V and 4V). When these capacitors are connected in parallel with reverse polarity, the charges on their plates redistribute in order to balance out the potential difference between them. This redistribution of charges results in the formation of a new electric field between the two capacitors, leading to the production of heat.

The amount of heat produced during this redistribution process can be calculated using the formula for energy stored in a capacitor, which is given by U = 0.5CV^2. In this case, we have two capacitors with different initial energies (U1 and U2) and different final energies (U1' and U2'). The difference between these two energies (U1'-U1 and U2'-U2) gives us the amount of heat produced during the redistribution process.

To calculate this heat, we need to consider the fact that the two capacitors have different capacitances, and hence, different amounts of charge will flow through them. This can be taken into account by using the formula Q = CV, where Q represents the charge on a capacitor, C represents its capacitance, and V represents the potential difference across it.

Using this formula, we can calculate the final charge on each capacitor after redistribution. For the capacitor with capacitance C, the final charge will be Q1 = CV1 + CV2, where V1 and V2 are the initial potential differences of the two capacitors. Similarly, for the capacitor with capacitance 2C, the final charge will be Q2 = 2CV1 + 2CV2.

Now, we can use the formula for energy stored in a capacitor to calculate the final energies of the two capacitors. U1' = 0.5CQ1^2 and U2' = 0.5(2C)Q2^2. The difference between these two energies will give us the amount of heat produced during the redistribution process.

Therefore, the final formula for calculating the heat produced during the redistribution of charges in a