Charge on capacitors each with a battery

In summary, the problem involves three uncharged capacitors, C1, C2, and C3, in a circuit with battery voltages V1=18 V, V2=10 V, V3=18 V and capacitances C1=8.9 μF, C2=20 μF, C3=12.2 μF. The goal is to determine the magnitude of the charge on each capacitor once equilibrium is established. To solve the problem, Kirchhoff's circuit law and the conservation of charge are used. By equating the potential gain/drop across each branch, three equations can be formed, but there are three unknowns. However, since the sum of charges on the plates must be zero
  • #1
reset_7
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Homework Statement


Three capacitors C1–C3, all initially uncharged, are placed in the circuit shown. The capacitances are C1=8.9 μF, C2=20 μF, C3=12.2 μF, and the battery voltages are V1=18 V, V2=10 V, V3=18 V.

What is the magnitude q1 of the charge on capacitor C1 once equilibrium has been established? Answer (given but I can't get it on my own): 0.000121 C

What is the magnitude q2 of the charge on capacitor C2 once equilibrium has been established? Answer (given but I can't get it on my own): 0.000287 C

What is the magnitude q3 of the charge on capacitor C3 once equilibrium has been established? Answer (given but I can't get it on my own): 0.000166 C

Homework Equations



Potential across each pair of battery / capacitor is the same.

The Attempt at a Solution



Equate the potential gain / drop from each branch to one another. But only get two equations (third is redundant) and have three unknowns.

Thanks in advance for any help.
 
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  • #2
Use conservation of charge.

ehild
 
  • #3
Hey,

What I did to work through this problem was to first determine which way the currents were going from the sources and then write an equation based on Kirchhoff's circuit law for the current distribution. From there you can rewrite the capacitors in their appropriate forms (series or parallel equivalents) and start solving for variables.

The one thing that had me caught up for quite some time was what the total voltage was... pay close attention to the circuit diagram.

Cheers
 
  • #4
ehild said:
Use conservation of charge.

ehild

Thanks for this message ehild, can you give me a bit more detail on the steps I would need to do in order to get the final solution?

Thanks again
 
  • #5
Rtjones said:
Hey,

What I did to work through this problem was to first determine which way the currents were going from the sources and then write an equation based on Kirchhoff's circuit law for the current distribution. From there you can rewrite the capacitors in their appropriate forms (series or parallel equivalents) and start solving for variables.

The one thing that had me caught up for quite some time was what the total voltage was... pay close attention to the circuit diagram.

Cheers

Thanks RTjones. I have tried this but am still not getting the answer. Did you get the final answers?

Thanks again for any further help you can provide.
 
  • #6
Yes I did get the correct answers. What I ended up doing to get the correct answer was using the EQn V = Q/Ceq. Where Ceq is the correct form for capacitance and Q is the total charge (which is subsequently the charge on capacitor 2). The part that was tripping me up though was that V is 28, not 18 or 10...the reason being because V1 and V2 are in series.

Using this V with the equation above you can solve for Qtotal (aka Q2) and then work your way back to the other charges via relating different paths to one another. Let me know if you're still struggling, I'd be glad to help...
 
  • #7
reset_7 said:
Thanks for this message ehild, can you give me a bit more detail on the steps I would need to do in order to get the final solution?

Thanks again

You can change the position of the third battery with capacitor 3, it will not alter the voltages. If the upper plates of all capacitors are connected, no charge can flow on them, so the sum of charges on the three plates is zero. On the plates, the individual charges can be either positive or negative. If the capacitors are charged to Q1, Q2, Q3, respectively, there is -Q1 charge on the right plate of C1, +Q2 on the upper plate of C2 and -Q3 on the upper plate of C3. See picture.

ehild
 
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  • #8
Rtjones said:
Yes I did get the correct answers. What I ended up doing to get the correct answer was using the EQn V = Q/Ceq. Where Ceq is the correct form for capacitance and Q is the total charge (which is subsequently the charge on capacitor 2). The part that was tripping me up though was that V is 28, not 18 or 10...the reason being because V1 and V2 are in series.

Using this V with the equation above you can solve for Qtotal (aka Q2) and then work your way back to the other charges via relating different paths to one another. Let me know if you're still struggling, I'd be glad to help...

Ok, thanks again for your help.
 
  • #9
ehild said:
You can change the position of the third battery with capacitor 3, it will not alter the voltages. If the upper plates of all capacitors are connected, no charge can flow on them, so the sum of charges on the three plates is zero. On the plates, the individual charges can be either positive or negative. If the capacitors are charged to Q1, Q2, Q3, respectively, there is -Q1 charge on the right plate of C1, +Q2 on the upper plate of C2 and -Q3 on the upper plate of C3. See picture.

ehild

Thanks, it was that fact that the charges must add up to zero due to plates being connecged that gave me my third equation.
 

What is the formula for calculating the charge on a capacitor?

The formula for calculating the charge on a capacitor is Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor in volts.

Can a capacitor have a charge without a battery?

Yes, a capacitor can have a charge without a battery. When a capacitor is connected to a voltage source, such as a battery, it stores electrical energy in the form of a charge on its plates. Once disconnected from the source, the capacitor can retain this charge.

How does the charge on a capacitor change when connected to a battery?

When a capacitor is connected to a battery, the voltage across the capacitor increases, which in turn increases the charge on the capacitor. This is because the battery supplies an electric field, causing the charges on the capacitor's plates to move and store energy.

What happens to the charge on a capacitor when the battery is removed?

When the battery is removed, the charge on the capacitor remains the same. The capacitor will continue to hold the charge until it is either discharged or connected to a different voltage source.

How does the charge on a capacitor affect its capacitance?

The charge on a capacitor does not affect its capacitance. The capacitance of a capacitor is a physical property determined by its size, shape, and material. However, the amount of charge on a capacitor is directly proportional to the capacitance, as shown by the formula Q = CV.

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