What will be the final voltage across the 2 micro-farad capacitor?

In summary: So your answer is right, I made a mistake.In summary, two capacitors of 1 and 2 micro-farad, charged by a 5 V battery and then connected to each other, will have a final voltage of -1.67 V across the 2 micro-farad capacitor. This is determined by the equation C1V1-C2V2=C1V'+C2V', where V1 and V2 represent the initial voltages of the capacitors, and V' represents the final voltage. The negative voltage results from the inversion of polarity in one of the capacitors, which does not affect the capacitance value but changes the sign of its charge.
  • #1
Krushnaraj Pandya
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Homework Statement


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?

Homework Equations


All applicable to capacitance

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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  • #2
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.
 
  • #3
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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  • #4
jbriggs444 said:
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
 
  • #5
Krushnaraj Pandya said:
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.
 
  • #6
Delta² said:
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?
 
  • #7
jbriggs444 said:
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?
 
  • #8
Krushnaraj Pandya said:
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.
 
  • #9
jbriggs444 said:
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.
 
  • #10
Delta² said:
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
 
  • #11
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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  • #12
Delta² said:
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)
 
  • #13
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.
 
  • #14
Delta² said:
-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.
 
  • #15
Ok let's 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##?
 
  • #16
Delta² said:
Ok let's 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)
 
  • #17
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##.
 
  • #18
Delta² said:
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
 
  • #19
Krushnaraj Pandya said:
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)
 
  • #20
Krushnaraj Pandya said:
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...
 
  • #21
To correct your reasoning, if the potential difference is x, then the positive charge is ##C|x|## and the negative charge ##-C|x|##..
 
  • #22
Delta² said:
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?
 
  • #23
Delta² said:
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?
 
  • #24
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.
 
  • #25
Delta² said:
The charges in those plates are both negative or both positive, after the equilibrium has been established
Always true?
 
  • #26
Krushnaraj Pandya said:
Always true?
Yes always true for two capacitors that have reached equilibrium. If we have 3 capacitors then it isn't necessarily true.
 
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  • #27
Delta² said:
Yes always true for two capacitors that have reached equilibrium. If we have 3 capacitors then it isn't necessarily true.
Alright, Thank you very much :D
 
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1. What is the formula for calculating final voltage across a capacitor?

The formula for calculating final voltage across a capacitor is V = Q/C, where V is voltage, Q is charge, and C is capacitance.

2. How is capacitance related to the final voltage across a capacitor?

Capacitance is directly proportional to the final voltage across a capacitor. This means that as the capacitance increases, the final voltage also increases, and vice versa.

3. What factors can affect the final voltage across a capacitor?

The final voltage across a capacitor can be affected by the amount of charge stored on the capacitor, the capacitance of the capacitor, and the initial voltage applied to the capacitor.

4. Why is the final voltage across a capacitor important in electronic circuits?

The final voltage across a capacitor is important in electronic circuits because it determines the amount of energy stored in the capacitor, which can then be used to power other components in the circuit.

5. Can the final voltage across a capacitor ever be greater than the initial voltage?

No, the final voltage across a capacitor cannot be greater than the initial voltage. The capacitor acts as a voltage divider, so the final voltage will always be less than or equal to the initial voltage.

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