I'm currently working on the following physics problem and have already explored various approaches to maximize the output. Below is the full task description, followed by a summary of my current thoughts and analysis.

In a box in the physics collection, you find a battery with voltage U, some wires, and two identical capacitors with capacitance C.
Find a method to use these components to achieve the highest possible voltage. Describe the steps you need to take and determine the maximum achievable voltage.
Hint: It is possible to achieve voltage values greater than 3 × U.

In working through this problem, I’ve considered both the standard and more advanced ways of increasing voltage with the given components.


A straightforward solution would be to charge the two capacitors in parallel (each to voltage U), and then reconfigure them in series together with the battery, which yields a total voltage of 3U across the output. This seems to be the obvious, textbook approach.


However, since the problem explicitly mentions that voltages greater than 3U are achievable, I’ve considered more sophisticated setups. One idea is to charge the two capacitors with opposite polarities — one to +U, the other to –U — and then combine them in series. By doing so, and perhaps integrating the battery in subsequent steps, it would be possible to gradually increase the voltage in steps, following the basic principle of a charge pump. This method would rely on repeated reconfiguration of the circuit to build up charge and transfer energy into a single direction.


The key uncertainty, however, lies in the limitations of the setup: only a battery, two capacitors, and wires are provided — no switches, no diodes. This raises the question of whether manual rewiring (i.e., physically changing the connections step by step) is considered an acceptable part of the task, or whether the mention of “greater than 3U” refers more to theoretical possibilities beyond what can be done with the strictly listed components.


Therefore, a crucial point of clarification is whether manual reconnection of the circuit is implicitly allowed or even expected in solving the problem — or if 3U is the practical maximum under the constraints of a static setup with no switching elements.

 

Should we assume that EMF and Voltage are considered interchangeable in this exorcise?

 

The problem explicitly asks for steps, plural. I take that to mean YOU can act as the switching device and the polarity controller.

IMHO, the challenge is to determine the maximum voltage the capacitors can handle, and that in turn limits the maximum voltage. Maybe I'm missing something.

 

Should we assume that EMF and Voltage are considered interchangeable in this exorcise?

 

Yes, I think in the context of this exercise, EMF and voltage can be treated as interchangeable.
Since the setup only involves ideal components (a battery, capacitors, and wires) and there’s no mention of internal resistance or energy losses, it seems reasonable to assume that the battery's EMF is equal to the voltage it provides across its terminals.


If this assumption were incorrect, the task would likely specify non-ideal behavior explicitly. So for the purposes of analysis, treating EMF as simply U should be valid.

 

The problem explicitly asks for steps, plural. I take that to mean YOU can act as the switching device and the polarity controller.

IMHO, the challenge is to determine the maximum voltage the capacitors can handle, and that in turn limits the maximum voltage. Maybe I'm missing something.

You're absolutely right — the wording "steps" (in plural) strongly suggests that manual intervention is allowed, i.e., we as solvers are expected to reconfigure the circuit actively, essentially acting as the switching mechanism and polarity controller ourselves.


That interpretation would make sense especially in light of the hint "voltages greater than 3U are possible," which can't be achieved in a static configuration. So yes, I agree: the exercise likely assumes that we are allowed (and even encouraged) to manually change the wiring between steps to simulate dynamic charge redistribution, similar to how a charge pump operates.


As for the second part of your message: you're also touching on a crucial physical limit. Even if we find a clever way to stack voltages through sequential charge transfers, the maximum voltage is ultimately limited by the breakdown voltage of the capacitors (i.e., how much electric field their dielectric can tolerate before failure). That could indeed cap how far this method can go in reality. But unless such a limit is specified in the problem, I’d assume the capacitors are ideal and can hold any voltage generated through the allowed process — which brings us back to a purely conceptual, ideal-physics setting.


So maybe the core challenge is:
Given that we can manually rewire the system step by step, what is the maximum voltage we can produce in principle, using only U, C, and symmetry?


Let me know if you're also working toward a generalized formula — I'd be interested in cross-checking the steps.

 

I'm currently working on the following physics problem and have already explored various approaches to maximize the output. Below is the full task description, followed by a summary of my current thoughts and analysis.





In working through this problem, I’ve considered both the standard and more advanced ways of increasing voltage with the given components.


A straightforward solution would be to charge the two capacitors in parallel (each to voltage U), and then reconfigure them in series together with the battery, which yields a total voltage of 3U across the output. This seems to be the obvious, textbook approach.


However, since the problem explicitly mentions that voltages greater than 3U are achievable, I’ve considered more sophisticated setups. One idea is to charge the two capacitors with opposite polarities — one to +U, the other to –U — and then combine them in series. By doing so, and perhaps integrating the battery in subsequent steps, it would be possible to gradually increase the voltage in steps, following the basic principle of a charge pump. This method would rely on repeated reconfiguration of the circuit to build up charge and transfer energy into a single direction.


The key uncertainty, however, lies in the limitations of the setup: only a battery, two capacitors, and wires are provided — no switches, no diodes. This raises the question of whether manual rewiring (i.e., physically changing the connections step by step) is considered an acceptable part of the task, or whether the mention of “greater than 3U” refers more to theoretical possibilities beyond what can be done with the strictly listed components.


Therefore, a crucial point of clarification is whether manual reconnection of the circuit is implicitly allowed or even expected in solving the problem — or if 3U is the practical maximum under the constraints of a static setup with no switching elements.

Hi there,

Yes it is possible.

If you charge two capacitors in parallel, then put them in series, you already have 2x the voltage U. However, you don't have to do that.
The question I ask you is this:
What if you charge just one cap and put it in series with the battery, then you have a source that is 2x the voltage of U ...
then what can you do with that second cap now that you have a voltage source that is 2x that of U?
There is at least one solution to get a voltage higher than 3*U but there might be two. See what you can find.

I don't know if you want to do this, but can you calculate the voltage(s) that would result (one or both greater than 3*U)?
This might be a little harder theoretically it's based on some careful physics.

 

If you charge the capacitors in parallel to the battery voltage, and then connect them all in series, there's no theoretical limit to the maximum voltage you can achieve.
It just depends upon the number of capacitors.

 

Hi there,

Yes it is possible.

If you charge two capacitors in parallel, then put them in series, you already have 2x the voltage U. However, you don't have to do that.
The question I ask you is this:
What if you charge just one cap and put it in series with the battery, then you have a source that is 2x the voltage of U ...
then what can you do with that second cap now that you have a voltage source that is 2x that of U?
There is at least one solution to get a voltage higher than 3*U but there might be two. See what you can find.

I don't know if you want to do this, but can you calculate the voltage(s) that would result (one or both greater than 3*U)?
This might be a little harder theoretically it's based on some careful physics.

That's a really interesting approach — and I appreciate how you've framed it as a series of logical steps, almost like a voltage "ladder" that can be climbed with clever wiring.


I do have a few follow-up questions though, just to make sure I fully understand what's happening here:

1. Why does the second capacitor (C2) charge up to 2U when connected across the battery + charged capacitor? I would have expected it to reach U, assuming some sort of redistribution — but the full 2U seems almost too "clean." Is that just a consequence of ideal components and fixed potentials?
2. Can this be repeated to reach even higher voltages — say, 5U or 6U — by alternating roles between the two capacitors in successive steps?
3. Does this violate energy conservation?
   If we're only using a battery with voltage U, how can we end up with capacitors charged to 2U, 3U, or 4U? I understand that the energy stored in capacitors scales with U², so it seems like the battery is doing more than it should be able to.
4. What ultimately limits the achievable voltage in this kind of system? Is it purely the breakdown voltage of the capacitors? Or could there also be circuit-level limits (e.g., how much charge they can hold, leakage, wire resistance, etc.)?
5. And finally: Is it really this simple, or is there something subtle we're overlooking? I feel like it works under ideal assumptions, but I wonder how robust this strategy would be in a real-world setup.

I’d love to hear your thoughts on this — and thanks again for sparking such a thought-provoking discussion.

 

If you charge the capacitors in parallel to the battery voltage, and then connect them all in series, there's no theoretical limit to the maximum voltage you can achieve.
It just depends upon the number of capacitors.

As stated earlier, there just two capacitors available, thats why it is so tricky

 

In the first stage of pumping, you place one charged capacitor at 1U in series with the battery. The combined voltage is 2U. You connect the second capacitor parallel to that. BUT the voltage on C1 will drop as the second cap pulls part of its charge. The final voltage on C2 is not 2U but something less, let's say 1.5U. The second pump step is to replace C1 with C2. Total voltage is now 2.5U (and NOT 3U). You can now put C1 in parallel first with the battery to get it to 1U and then in parallel with the battery and C2 in series. The C1 voltage would increase from 1U up to something less than 2.5U, and so on.

If you work out the math and solve for voltage versus number of steps, I think you'll find the answer converges fairly quickly. Or just plot actual data of voltage versus pump cycles. It doesn't go to infinity but tops out. It doesn't need the limit I was thinking of earlier, the capacitor's breakdown voltage. I agree you can probably assume an ideal capacitor that doesn't break down.

 

Connect both capacitors in parallel across the battery. Each will charge up to the battery voltage.
Then connect one capacitor in series with the battery across the second capacitor.
Then recharge the first capacitor with the battery.
Then connect it in series with the battery across the second capacitor.
If you repeat the last two steps an infinite number of times, the second capacitor will charge up to twice the battery voltage.
Connecting both capacitors in series with the battery then will give a total voltage of four times the battery voltage.

 

Connect both capacitors in parallel across the battery. Each will charge up to the battery voltage.
Then connect one capacitor in series with the battery across the second capacitor.
Then recharge the first capacitor with the battery.
Then connect it in series with the battery across the second capacitor.
If you repeat the last two steps an infinite number of times, the second capacitor will charge up to twice the battery voltage.
Connecting both capacitors in series with the battery then will give a total voltage of four times the battery voltage.

Hey,

Im thankful for your response! Can you lay out the physics behind your idea. How excatly (or to put it differntly why) does it work up to 4U? Can you explain it in more detail?

Best regards

 

In the first stage of pumping, you place one charged capacitor at 1U in series with the battery. The combined voltage is 2U. You connect the second capacitor parallel to that. BUT the voltage on C1 will drop as the second cap pulls part of its charge. The final voltage on C2 is not 2U but something less, let's say 1.5U. The second pump step is to replace C1 with C2. Total voltage is now 2.5U (and NOT 3U). You can now put C1 in parallel first with the battery to get it to 1U and then in parallel with the battery and C2 in series. The C1 voltage would increase from 1U up to something less than 2.5U, and so on.

If you work out the math and solve for voltage versus number of steps, I think you'll find the answer converges fairly quickly. Or just plot actual data of voltage versus pump cycles. It doesn't go to infinity but tops out. It doesn't need the limit I was thinking of earlier, the capacitor's breakdown voltage. I agree you can probably assume an ideal capacitor that doesn't break down.

thanks again for your helpful response earlier – it clarified a lot about how the voltage progression behaves with just two capacitors and a single battery.

I’ve been trying to work through the math behind this step-by-step, but I’m still a bit unsure how exactly to calculate the voltage over time (or per pump cycle). Would you mind helping me work through a few steps of the actual calculations? Even just the recursive formula or a short example would help me understand the underlying behavior much better.

Of course, only if you have the time – I’d really appreciate it!

 

Well, since the cat has been let out of the bag.

Charge both caps to 1U.
Connect them in a "complete" series circuit, having C1's negative terminal connected to the battery positive and C2 in "backwards".
Disassemble the circuit and recharge C1 to 1U.
Connect all three back into an "open" series circuit with the negatives connected to the positives.

Bingo, you should now have a voltage above 3*U.

In a nutshell, you are simply using the battery and C1 to raise the voltage on C2.

Please don't ask me to do any math.

 

I believe our TS needs to do either the math or an actual experiment, or both, to receive credit.

I suggest starting with the situation of the battery plus one capacitor C1, total voltage 2U, and then attach the loose capacitor C2, already at 1U, in parallel. Where does the voltage on C2 end up? In other words, how does the charge distribute?

 

I believe our TS needs to do either the math or an actual experiment, or both, to receive credit.

I suggest starting with the situation of the battery plus one capacitor C1, total voltage 2U, and then attach the loose capacitor C2, already at 1U, in parallel. Where does the voltage on C2 end up? In other words, how does the charge distribute?

I agree! We have explained how to do it. Now it's up to him to figure out the details.

 

Connect both capacitors in parallel across the battery. Each will charge up to the battery voltage.
Then connect one capacitor in series with the battery across the second capacitor.
Then recharge the first capacitor with the battery.
Then connect it in series with the battery across the second capacitor.
If you repeat the last two steps an infinite number of times, the second capacitor will charge up to twice the battery voltage.
Connecting both capacitors in series with the battery then will give a total voltage of four times the battery voltage.

MOD NOTE: Ah... Homework Help, not Homework Done For You.

 

it would be possible to gradually increase the voltage in steps, following the basic principle of a charge pump. This method would rely on repeated reconfiguration of the circuit to build up charge and transfer energy into a single direction.

The key uncertainty, however, lies in the limitations of the setup: only a battery, two capacitors, and wires are provided — no switches, no diodes. This raises the question of whether manual rewiring (i.e., physically changing the connections step by step) is considered an acceptable part of the task, or whether the mention of “greater than 3U” refers more to theoretical possibilities beyond what can be done with the strictly listed components.

But the "obvious" way that gets you 3*U also requires manual rewiring, so manual rewiring must be allowed at least once during the process.

You mentioned following the basic principle of a charge pump, so as yourself what role the diodes and input waveform serve and how to replicate them via manual manipulations of the circuit wiring.

Therefore, a crucial point of clarification is whether manual reconnection of the circuit is implicitly allowed or even expected in solving the problem — or if 3U is the practical maximum under the constraints of a static setup with no switching elements.

To the degree that your question is about what is allowed or expected in solving the problem, that is a question for the person that assigned the problem, not strangers on the Internet. The best we can do here is guess at what makes sense, with no guarantee that we will guess correctly (i.e., successfully read the mind of the person that wrote, assigned, and/or will grade the problem). So, even if you think our crystal balls seem to be working well on this one, it would be a good idea to touch base with the instructor. Not only will you get the answer that matters, as far as your grade is concerned, but it also gives you the opportunity to demonstrate to your instructor that you are thinking about the fine details of the problem, something that is all-too-often in short supply these days.

 

I believe our TS needs to do either the math or an actual experiment, or both, to receive credit.

I suggest starting with the situation of the battery plus one capacitor C1, total voltage 2U, and then attach the loose capacitor C2, already at 1U, in parallel. Where does the voltage on C2 end up? In other words, how does the charge distribute?

Hi again everyone,


Thanks to our previous conversation, I’ve been thinking more deeply about the charge pump setup we discussed – where two identical capacitors are alternately charged and connected in different configurations to gradually increase the output voltage.


Here’s the idea as I’ve understood and explored further (thanks to your input):

1. Step 1: Capacitor C₁ is charged to the battery voltage U.
2. Step 2: C₁ is placed in series with the battery (total ≈ 2U), and an initially uncharged capacitor C₂ is connected in parallel.
3. Step 3: C₂ charges up to some fraction of the total voltage due to charge redistribution – for instance, ≈4/3(a)⋅2U=1.5U\approx \frac{3}{4} \cdot 2U = 1.5U≈4/3⋅2U=1.5U.
4. Step 4: The cycle repeats with roles reversed: C₂ now acts in series, and C₁ is charged again.

Each cycle transfers less charge than the previous one, so the voltage increases asymptotically and never reaches a strict upper limit. But the final attainable voltage depends heavily on the transfer efficiency factor aaa.


What I’ve explored based on your idea:

• I tried modeling this using
  
  
  as the asymptotic voltage for one capacitor.
• 
  
• With a= 4/3
  G=3U(per capacitor + battery )⇒up to 3U total in series with battery.
• If we use a different term for a it would significantly lead us to a different max voltage

What I’m now wondering:

Is there any simulation, literature, or analytic reasoning that supports using a=4/3 as a realistic approximation?


Or is this just an optimistic estimate?


I understand that some energy is always lost due to redistribution (even in ideal models), and that the actual value of a would depend on how charge equilibrates in each step – but I’m not sure what value is most accurate.


Thanks again for your insights – you really helped me get this far!