# Final charges on the capacitors

1. Jun 20, 2012

1. The problem statement, all variables and given/known data

The Problem is as follows :-

https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-snc6/224892_2277870043526_218311126_n.jpg [Broken]

Three capacitors of 2 μF, 3μF, and 5μF are indulgently charged with batteries of emf’s 5V,20V and 10V respectively. After disconnected from the batteries these capacitors are joined as shown in the figure with their positive polarity plates are connected to A and negative polarity plates are earthed. Now a battery of 20 V and an uncharged capacitor of 4μF are joined to A as shown with the switch S. When the switch is closed find (i) The potential of junction A, (ii) Final charges on all four capacitors.

2. Relevant equations & The attempt at a solution

as the question says Initially the charges on 2 μF, 3μF, and 5μF are 10 μc, 60 μc, 50 μc respectively. After the connection to ‘A’ charge will flow in all the capacitor in such a manner to make the potential of connected plates same. For to do this I used the formula of common potential V_C=(C_1 V_1+〖 C〗_2 V_2+〖 C〗_3 V_3)/(〖 C〗_1+〖 C〗_2+〖 C〗_3 )
After using this formula I got the potential of Point ‘A’ as 12 V.
Now a 20 V battery (+)ve terminal is connected to point A hence the potential should be 20 + 12 = 32V.
But the answers are as follows:
(i) 100/7 V
(ii) Charges - 28.56 μc, 42.84 μc, 71.4 μc, 22.88 μc

Last edited by a moderator: May 6, 2017
2. Jun 20, 2012

### Ratch

No it's not. The 4uf cap energizes to a minus voltage, so that has to be subtracted from the 20 volts. I got the correct answers by solving a differential equation. Maybe someone else knows a simpler way.

Ratch

3. Jun 20, 2012

Dear friend Could you please tell me the way to solve the problem by differential equation?
I'll be very thankful to you for this help.

4. Jun 25, 2012

### Yukoel

Hello thunderharon,
Point A and battery are connected by a conducting wire right ?So this indicates a mistake in your common potential combination.
This problem can be handled in without differential equations as well.
C(2μF),C(3μF) and C(5μF) have initial charges as 10,60 and 50 μC respectively.It is given that they are connected with their positive plates to A.
After the switch is closed assume that the battery gives them surplus charges of q1,q2,q3 and q4 respectively.So that the new charges are 10+q1,60+q2,50+q3 and q4 respectively.Now since battery supplies equal amount of charge through both its ends (meaning it supplies as much positive charge through positive end as negative from the negative end). As such
q1+q2+q3=q4 $1$
Now let us start from the earth to reach Point A through different branches.Potential of earth is taken as zero.
V(a)=(10+q1)/2 (from 2μF branch)=(60+q2)/3(from 3μF branch)=(50+q3)/5(from 5μF branch)=20-(q4/4) [from the branch containing 4μF and battery] $2$
Solve 1 and 2 to get the desired result.
regards
Yukoel

5. Jun 25, 2012

Hi Yukoel !

I solve the equations and got the correct answer. Thank you very much for your outstanding reply. The problem is solved.

In your post you wrote "surplus charges". Can you please make me understand this term.

6. Jun 25, 2012

### Yukoel

Well I had to define a new set of charges due to the battery; all I meant to say was that the battery let out a certain amount of positive charge which split and was stored as q1, q2 and q3 on the plates.The same amount of negative charge i.e. q4 was stored on the uncharged capacitor.So in short I loosely used the term surplus charge to get the new charges on the capacitor .It is not a technical term.But without it the equations could have been a little uglier . :)
regards
Yukoel

7. Jun 25, 2012