# Connecting two charged capacitors in series

• vcsharp2003
In summary, we have two unknown charges, ##q_3'## and ##q_4'##, after a state of balance is reached. The relationship between these charges is given by the equation ##q_4-q_3=q_3'+q_4'##, which is based on the law of charge conservation. This equation is derived from the fact that the two capacitors are connected in reverse polarity, meaning that the left plates of each capacitor are interconnected. Thus, applying charge conservation for the left pair of plates, we get the equation ##q_4-q_3=-(q_3'+q_4')##, which is equivalent to the first equation. Solving these equations will give us the values of

#### vcsharp2003

Homework Statement
Two capacitors of ##3 \text{ } \mu \text{F}## and ##4 \text{ } \mu \text{F}## are connected so negative plate of one is attached to positive plate of the other as shown in diagram. What is the final charge on each capacitor? These capacitors were initially charged using a 6 V battery.
Relevant Equations
C = V /q

Also, charges should be conserved since new charges cannot be created or existing ones destroyed
I looked at the solution of this problem since its a solved problem. I am having doubts with the charges relationship as is mentioned in screenshot below. The charges ##{q_3}^{'}## and ##{q_4}^{'}## are the charges after a a state of balance is reached.

Why would the charges have the relationship as mentioned in screenshot below? I have determined ## q_3 = 18 \text{ } \mu \text{F}## and ## q_4 = 24 \text{ } \mu \text{F}## using the equation q = CV. Last edited:
by kirchhoffs joint rule or whatever it's called. Some charge , let's say q travels from negative plate of 4uF and gets deposited at positive plate of 3uF capacitor, by induction current through plates of capacitor, q is also passed onto negative plate of the 3uF capacitor, flowing back to positive plate of 4uF capacitor
Now in this situation, current has flowed to clear the voltage difference ,so as the capacitors have differnt charges which result in zero potential difference, according to terminology of question
4uF capacitor has a charge q4 +q=q4'
and 3uF capacitor has a charge -(q3+q)=q3'
so if we add above equations then q4'+q3'=q4-q3
hope you are able to understand my approach

• vcsharp2003
bhupesh said:
by kirchhoffs joint rule or whatever it's called. Some charge , let's say q travels from negative plate of 4uF and gets deposited at positive plate of 3uF capacitor, by induction current through plates of capacitor, q is also passed onto negative plate of the 3uF capacitor, flowing back to positive plate of 4uF capacitor
Now in this situation, current has flowed to clear the voltage difference ,so as the capacitors have differnt charges which result in zero potential difference, according to terminology of question
4uF capacitor has a charge q4 +q=q4'
and 3uF capacitor has a charge -(q3+q)=q3'
so if we add above equations then q4'+q3'=q4-q3
hope you are able to understand my approach

Is the charge q that you mentioned a +ve or -ve charge?

i have not given any polarity to the charge q

bhupesh said:
i have not given any polarity to the charge q

But if charge q flows from -ve plate to +ve plate, then shouldn't some negative charge be flowing from the -ve plate since it has excess of negative charges (i.e. initially -24 charge on -ve plate and +18 charge on +ve plate).

charge flows due to potential difference not due to some excess
and I clearly implied that i have not mentioned the polarity of q

bhupesh said:
charge flows due to potential difference not due to some excess
and I clearly implied that i have not mentioned the polarity of q

It doesn't make sense to me. If potential difference makes a charge q flow, then if we take convention of current a positive charge must be flowing.

Also how can charge be deposited, but it is neither +ve or -ve?

bhupesh said:
4uF capacitor has a charge q4 +q=q4'
and 3uF capacitor has a charge -(q3+q)=q3'

Why did you put a negative sign for the 3uF capacitor?

OMG. There is a picture in #1 which defines the ##q_3'~and~q_4'##. These are unknown. Don't define anything else. Write down two equations (charge conservation and capacitance/ voltage). Solve for the two unknowns.

• alan123hk
hutchphd said:
OMG. There is a picture in #1 which defines the ##q_3'~and~q_4'##. These are unknown. Don't define anything else. Write down two equations (charge conservation and capacitance/ voltage). Solve for the two unknowns.
I think the OP question is why we take conservation of charge equation as ##q_3'+q_4'=q_4-q_3## and not as ##q_3'+q_4'=q_4+q_3##. I am sure the answer is "because we connect the capacitors with reverse polarity" but the OP seeks for a more detailed answer.

• vcsharp2003
The way I understand it we must be more explicit on how we apply charge conservation: We apply it for the right pair of the plates or for the left pair of the plates, because those pairs of plates are interconnected. So if we apply it for the left pair of plates:
BEFORE: left plate of the up capacitor has charge ##q_3## (positive) while left plate of the down capacitor has charge ##(-q_4)## (negative). Total charge ##q_3-q_4##
AFTER: left plate of the up capacitor has charge ##-q_3'## (negative) and left plate of the down capacitor has charge ##(-q_4')## (negative). Total charge ##-q_3'-q_4'##
So the equation BEFORE=AFTER becomes ##q_3-q_4=-(q_3'+q_4')## or equivalent ##q_4-q_3=q_3'+q_4'##.

• alan123hk and vcsharp2003
Delta2 said:
The way I understand it we must be more explicit on how we apply charge conservation: We apply it for the right pair of the plates or for the left pair of the plates, because those pairs of plates are interconnected. So if we apply it for the left pair of plates:
BEFORE: left plate of the up capacitor has charge ##q_3## (positive) while left plate of the down capacitor has charge ##(-q_4)## (negative). Total charge ##q_3-q_4##
AFTER: left plate of the up capacitor has charge ##-q_3'## (negative) and left plate of the down capacitor has charge ##(-q_4')## (negative). Total charge ##-q_3'-q_4'##
So the equation BEFORE=AFTER becomes ##q_3-q_4=-(q_3'+q_4')## or equivalent ##q_4-q_3=q_3'+q_4'##.

That makes a lot of sense. But I have one doubt and that is how can we conclude that when -ve left plate of down capacitor is connected to +ve left plate of up capacitor, then both these plates end up with -ve charge finally?

• alan123hk
We chose a clear definition of what the charges are and the net charge on the system starts at zero. This allows us to determine uniquely the numbers.
I believe your confusion comes from the overall lack of uniqueness. We could uniformly increase the net charge on the system without changing the physics at all. Only the separated charge and the potential (Voltage) differences matter within the system. If the net charge is nonzero interaction with the rest of the world may matter. But with the conditions given the result is what it is.

Edit: These capacitors are connected in parallel.

• alan123hk
vcsharp2003 said:
That makes a lot of sense. But I have one doubt and that is how can we conclude that when -ve left plate of down capacitor is connected to +ve left plate of up capacitor, then both these plates end up with -ve charge finally?
Well we don't know it a priori, we have to do the math first. But then how do we right down the charge conservation equation? We would have to write it down as ##q_3-q_4=q_3'+q_4'## and since we don't know whether ##q_3',q_4'## are positive or negative, when we form the equation for the voltages we would have to put them inside absolute values i.e. $$\frac{|q_3'|}{C_3}=\frac{|q_4'|}{C_4}$$ and when solving the system of equations we would have to take cases for ##q_3'## and ##q_4'##.

• vcsharp2003
Actually we have to take only the cases where ##q_3',q_4'## have the same sign, the other cases are not valid because it would mean that the capacitors have opposite polarities even after the process has ended.

• vcsharp2003 and alan123hk
Delta2 said:
Actually we have to take only the cases where q3′,q4′ have the same sign, the other cases are not valid because it would mean that the capacitors have opposite polarities even after the process has ended

This analysis is very good.

Because two capacitors connected in parallel must have the same voltage, their charges must have the same sign.

Since q3'and q4' must have the same sign, at the moment when the two capacitors are connected in parallel, current must flow from the left plate of the q3 capacitor to the left plate of the q4 capacitor.

hutchphd said:
OMG. There is a picture in #1 which defines the q3′ and q4′. These are unknown. Don't define anything else. Write down two equations (charge conservation and capacitance/ voltage). Solve for the two unknowns.

I agree with this. I suggest not to think too complicated for the time being. First, write down the equations for the system according to the definition of capacitor ##~VC=Q ~## , and then solve them.

Delta2 said:
Well we don't know it a priori, we have to do the math first. But then how do we right down the charge conservation equation? We would have to write it down as ##q_3-q_4=q_3'+q_4'## and since we don't know whether ##q_3',q_4'## are positive or negative, when we form the equation for the voltages we would have to put them inside absolute values i.e. $$\frac{|q_3'|}{C_3}=\frac{|q_4'|}{C_4}$$ and when solving the system of equations we would have to take cases for ##q_3'## and ##q_4'##.

According to the logic you mentioned, we end up with the following two equations. Equation 1 is obtained by applying conservation of charge rule to left plates of up and down capacitors. Equation 2 is obtained by considering that at equilibrium the voltage drop across top and down capacitors must be equal else electrons flow would be there (V = Q/C for a capacitor).

##+18-24=q_3'+q_4' \text { ....(1)}##
##\dfrac{|q_3'|}{3}=\dfrac{|q_4'|}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text { ....(2)} ##

From equation 1 we can say that ##q_3' = -6 - q_4'##. We substitute this into equation 2 to get the following.
##\dfrac{| -6 - q_4'|}{3}=\dfrac{|q_4'|}{4} \text { ....(2)} ##

We can use a simple mathematical trick here which says that if ##|{a}| = |{b}|## then ## a =\pm b## to reduce equation 2 to the following form.
##\dfrac{(-6 - q_4')}{3}= \pm \dfrac{q_4'}{4} \text { ....(2)} ##

Solving above equation 2 we get following two sets of solutions.

CASE 1:
##q_4' =-\dfrac {24} {7} = -3.43 \, \mu \text{C} \text { and } q_3' =-\dfrac {14} {7} = -2.57 \, \mu \text{C}##

Case 2:
##q_4' =-24 \, \mu \text{C} \text { and } q_3' =18 \, \mu \text{C}##

The answer is the values in Case 1, which also tells us that the left plates of up and down capacitors become negatively charged.
My question now is why would Case 2 be ignored? May be because the values are same as before the capacitors are connected or some other reason.

Last edited:
• Delta2
Case 2 is ignored because the charges are not of the same sign

• vcsharp2003
Delta2 said:
Case 2 is ignored because the charges are not of the same sign

Also, we could ignore it because some electrons will always be transferred from -ve left plate of down capacitor to +ve left plate of top capacitor which would change the charge values on both the left plates. Is that right?

Yes, that's the process I talk about in post #15. Once this process start it will end with same sign charge in the connected plates, otherwise the connected plates won't be on equal potential. One plate would have negative potential and the other would have positive (according to their charges that is).

Last edited:
• vcsharp2003
Delta2 said:
Yes, that's the process I talk about in post #15. Once this process start it will end with same sign charge in the connected plates, otherwise the connected plates won't be on equal potential. One plate would have negative potential and the other would have positive (according to their charges that is).

I find your approach is very precise and rigorous because it assumes even the signs of final charges on left plates as unknowns, and also the way you explained how charge conservation is applied is so much to the point. Top quality approach and as a result I fully understand the concepts now. Thankyou for your help.

• Delta2
vcsharp2003 said:
I find your approach is very precise and rigorous because it assumes even the signs of final charges on left plates as unknowns, and also the way you explained how charge conservation is applied is so much to the point. Top quality approach and as a result I fully understand the concepts now. Thankyou for your help.
Thank you, these words are the best reward for me.

• vcsharp2003