# Capacitor system problem

#### Krushnaraj Pandya

Gold Member
1. The problem statement, all variables and given/known data
Two capacitors of 1 and 2 micro-farad respectively are each charged by being connected across a 5 V battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2 micro-farad capacitor?

2. Relevant equations
All applicable to capacitance

3. The attempt at a solution
First I though that after being charged, both have a potential 5 V...connecting them shouldn't change anything but then I realized that since E is inverted for one capacitor because of connecting reverse polarities, V will be -5 V for one and 5 V for the other (I assumed 1 mF capacitor to have 5 V) then conserving charge for positive plate of 1 mF and negative plate of 2 mF capacitor, I wrote C1V1-C2v2=C1V-C2V replacing; 5-(-10)=1V-2V, so -15=V which is an absurd result. Where am I going wrong, I'd appreciate some help

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#### jbriggs444

Science Advisor
Homework Helper
It is a sign convention problem.

In effect, you have given one capacitor a negative capacitance. Rather than giving it a negative capacitance, just give it a negative charge.

#### Delta2

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Gold Member
if V, V' are the voltages before and after the interconnection , then the correct equation is $C_1V-C_2V=C_1V'+C_2V'$ because after the equilibrium has established and we have voltage V', each capacitor has the same sign in his charge at that specific plate to which we applied conservation of charge. The charges in those plates are both negative or both positive, after the equilibrium has been established.

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#### Krushnaraj Pandya

Gold Member
It is a sign convention problem.

In effect, you have given one capacitor a negative capacitance. Rather than giving it a negative capacitance, just give it a negative charge.
I have given it a negative potential difference and positive charge, since the plates are in reverse polarity-field and thus potential are effectively inverted. I don't understand the logic behind giving it a negative charge

#### jbriggs444

Science Advisor
Homework Helper
I have given it a negative potential difference and positive charge, since the plates are in reverse polarity-field and thus potential are effectively inverted. I don't understand the logic behind giving it a negative charge
A capacitor turned end for end has opposite charge but the same capacitance it started with.

#### Krushnaraj Pandya

Gold Member
if V, V' are the voltages before and after the interconnection , then the correct equation is $C_1V-C_2V=C_1V'+C_2V'$ because after the equilibrium has established and we have voltage V', each capacitor has the same sign in his charge at that specific plate to which we applied conservation of charge. The charges in those plates are both negative or both positive, after the equilibrium has been established.
Is this universally true for any system of capacitors?

#### Krushnaraj Pandya

Gold Member
A capacitor turned end for end has opposite charge but the same capacitance it started with.
Logic behind this?
And what's wrong with my logic?

#### jbriggs444

Science Advisor
Homework Helper
Logic behind this?
A[n ideal] capacitor is a pair of plates with a dielectric between. The two sides are identical -- mirror images. Inverting it changes nothing about its physical properties.

But if you swap the ends, any positive charge on one plate swaps places with any negative charge on the other plate. You've negated its charge and, accordingly, the potential difference between its ends. You've negated voltage and charge, but not capacitance.

#### Krushnaraj Pandya

Gold Member
A[n ideal] capacitor is a pair of plates with a dielectric between. The two sides are identical -- mirror images. Inverting it changes nothing about its physical properties.

But if you swap the ends, any positive charge on one plate swaps places with any negative charge on the other plate. You've negated its charge and, accordingly, the potential difference between its ends. You've negated voltage and charge, but not capacitance.
Alright. Got it, I'll apply this and try to reach the answer.

#### Krushnaraj Pandya

Gold Member
if V, V' are the voltages before and after the interconnection , then the correct equation is $C_1V-C_2V=C_1V'+C_2V'$ because after the equilibrium has established and we have voltage V', each capacitor has the same sign in his charge at that specific plate to which we applied conservation of charge. The charges in those plates are both negative or both positive, after the equilibrium has been established.
The equation you wrote still gives an incorrect answer. 5-(-10)=1V+2V, so V=5, the answer is 1.7 V

#### Delta2

Homework Helper
Gold Member
Ehm I wonder how you apply the equation, the way I apply it is $1x5-2x5=1V+2V \Rightarrow -5=3V \Rightarrow V=-5/3$, the minus sign is just a convention here.

You don't have to negate the voltage, I already took the minus sign in front so I wrote $C_1V-C_2V$ where the minus sign indicates that the charge in the second capacitor is viewed as negative...That is I applied conservation of charge to the negative plate of $C_2$ and the positive plate of $C_1$...

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#### Krushnaraj Pandya

Gold Member
Ehm I wonder how you apply the equation, the way I apply it is 5-10=1V+2V.. -5=3V, V=-5/3, the minus sign is just a convention here.

You don't have to negate the voltage, I already took the minus sign in front so I wrote $C_1V-C_2V$ where the minus sign indicates that the charge in the second capacitor is viewed as negative...That is I applied conservation of charge to the negative plate of $C_2$ and the positive plate of $C_1$...
Charge is definitely negative so we wrote -C2V instead of +C2V, but we also took V for the 2 mF capacitor to be -5 initially (if we took both as 5 V then no change should happen)

#### Delta2

Homework Helper
Gold Member
Read the equation again please, it is $C_1V-C_2V$ it is NOT $C_1V-C_2(-V)$… if we take as V=-5Volt then use of this is fine in the equation, just don't use -5Volt for one capacitor and +5Volt for the other capacitor, that's NOT correct.

#### Krushnaraj Pandya

Gold Member
-5Volt for one capacitor and +5Volt for the other capacitor, that's NOT correct
This is what I am doing, I am thinking that both have V of opposite signs otherwise no change is possible (if they already have same potential i.e both +5 or -5 V) logically supported by the fact that both have charged plates in opposite directions.

#### Delta2

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Gold Member
Ok lets say that the capacitors have opposite voltages, $C_1$ has $V$ and the $C_2$ has $-V$. If the positive charge in capacitor $C_1$ is $C_1V$, what is the negative charge in the capacitor $C_2$?

#### Krushnaraj Pandya

Gold Member
Ok lets say that the capacitors have opposite voltages, $C_1$ has $V$ and the $C_2$ has $-V$. If the positive charge in capacitor $C_1$ is $C_1V$, what is the negative charge in the capacitor $C_2$?
positive charge on c2 is C2(-V) therefore negative charge on it is also C2(-V) in magnitude. The expression should be C1V-C2(-V) (-ve sign outside since negative charge)

#### Delta2

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Gold Member
I assumed V to be positive (I said positive charge in $C_1$ is $C_1V$ which had to be a positive number). So you saying that positive charge on $C_2$ is $C_2(-V)$ which is a negative number. This is actually the negative charge on $C_2$.

#### Krushnaraj Pandya

Gold Member
I assumed V to be positive. So you saying that positive charge on $C_2$ is $C_2(-V)$ which is a negative number. This is actually the negative charge on $C_2$.
I assumed V to be +5 as well. But we are taking pot. = -V for C2. Therefore positive charge=C2(-V). What's wrong with this?
P.S- I have a feeling that there'll be an extremely silly mistake that I'm overlooking, but I would still like to find it so I can take care to not make it again

#### Krushnaraj Pandya

Gold Member
I assumed V to be +5 as well. But we are taking pot. = -V for C2. Therefore positive charge=C2(-V). What's wrong with this?
P.S- I have a feeling that there'll be an extremely silly mistake that I'm overlooking, but I would still like to find it so I can take care to not make it again
To make my thought process even clearer. When a capacitor has potential difference x across its plates, positive charge in it is Cx. Since on C2 , x=(-V), positive charge is C2(-V)

#### Delta2

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Gold Member
To make my thought process even clearer. When a capacitor has potential difference x across its plates, positive charge in it is Cx. Since on C2 , x=(-V), positive charge is C2(-V)
Positive charge can NOT be a negative number... If we follow your reasoning it is clear to me that $C_2(-V)$ which you consider to be the positive charge, is a negative number...

#### Delta2

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Gold Member
To correct your reasoning, if the potential difference is x, then the positive charge is $C|x|$ and the negative charge $-C|x|$..

#### Krushnaraj Pandya

Gold Member
Positive charge can NOT be a negative number... If we follow your reasoning it is clear to me that $C_2(-V)$ which you consider to be the positive charge, is a negative number...
Alright, this makes it clear that my reasoning is wrong.But where? how is post #19 wrong?

#### Krushnaraj Pandya

Gold Member
To correct your reasoning, if the potential difference is x, then the positive charge is $C|x|$ and the negative charge $-C|x|$..
oh ok. Why is this never mentioned anywhere?

#### Delta2

Homework Helper
Gold Member
Because usually we take the voltage of the capacitor as positive, so the positive charge is simply Q=CV. But if you consider the voltage of the capacitor as negative then you have to do as I say in post #21.

#### Krushnaraj Pandya

Gold Member
The charges in those plates are both negative or both positive, after the equilibrium has been established
Always true?

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