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[/color]
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.