A challenging capacitor problem

  • #1
1. Two identical flat plate capacitors are inserted in each other. First, none of the plates was charged, but then they have been connected to sources of current, keeping the constant voltage ## V_1 ## and ##V_2 ##. Find the potential difference between inner plates, which are kept at the distance a one to another. The distance between the capacitor plates is d.

2. ## -gradφ=E ##
## E=\frac σ ε_0 ##
##∫Edl=0## (over a closed loop)


3. I tried to solve this problem in many ways. For instance, I tried writing the electric field intensity E and making use of the circulation theorem, supposing that the plates are mutually inducing the same surface charge distribution σ, but I got a weird result. I tried to write the potential of each plate as electric field intensity multiplied by the distance between the plates, but I did not get a satisfying result. I am clueless what to do next and I don't know what is wrong in my approach.
 

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  • #2
kuruman
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... then they have been connected to sources of current, ...
You mean "sources of constant voltage", no?

I am not sure you should be worried about induced charges. The job of a voltage source is to keep the potential difference (voltage) across its terminals constant and the plates are equipotentials. What is the potential difference between the two plates at the very bottom that are separated by distance ##d-a##?
 
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  • #3
You mean "sources of constant voltage", no?

I am not sure you should be worried about induced charges. The job of a voltage source is to keep the potential difference (voltage) across its terminals constant and the plates are equipotentials. What is the potential difference between the two plates at the very bottom that are separated by distance ##d-a##?
If I suppose that the third and the fourth plate have a different surface charge distribution, then ##Δφ = \frac {(σ_4+σ_3)(d-a)} { 2 ε_0} ##
 
  • #4
SammyS
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Homework Statement


. Two identical flat plate capacitors are inserted in each other. First, none of the plates was charged, but then they have been connected to sources of current, keeping the constant voltage ## V_1 ## and ##V_2 ##. Find the potential difference between inner plates, which are kept at the distance a one to another. The distance between the capacitor plates is d.

Homework Equations


## -grad φ=E ##
## E=\frac σ ε_0 ##
##∫Edl=0## (over a closed loop)

The Attempt at a Solution


I tried to solve this problem in many ways. For instance, I tried writing the electric field intensity E and making use of the circulation theorem, supposing that the plates are mutually inducing the same surface charge distribution σ, but I got a weird result. I tried to write the potential of each plate as electric field intensity multiplied by the distance between the plates, but I did not get a satisfying result. I am clueless what to do next and I don't know what is wrong in my approach.
In the future, please leave the template headings in the opening Post of a Thread. (I put them back in the quoted text immediately above.)

Thanks for uploading your image. It's helpful to see the full-size image. Here's yours:
51927932_2252710261614064_1539607722482204672_n.jpg

Click the FULL IMAGE button when you upload an image.

I also applaud you for using LaTeX .
Here are a few hints to improve the LaTeX you displayed.

The " \ " (backslash) is the LaTeX escape character. Those Greek characters are best obtained by using \sigma , \varphi and \varepsilon .

To include more than one character in a numerator, denominator, subscript or superscript, enclose the desired characters in braces { } : as in \frac \sigma {\varepsilon_0} .

I have repeated (some of) your equations below with the LaTeX code exposed for each.

## -grad φ=E ##
## - \nabla \varphi = \vec E \ \quad \quad ## ## - \nabla \varphi = \vec E ##
## E=\frac σ ε_0 ##
## E = \frac \sigma {\varepsilon_0} \quad \quad ## ## E = \frac \sigma {\varepsilon_0} ##​
##∫Edl=0## (over a closed loop)
## \oint \vec E \cdot d \vec{l} = 0 \quad \quad ## ## \oint \vec E \cdot d \vec{l} = 0 ##​
 

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  • #5
collinsmark
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I'm not totally sure I understand the configuration. But the way I presently understand it, the physical configuration is as in the figure below,

Merged_caps.png


where [itex] V_a [/itex] is the electric potential that we are looking for. But I've pretty much convinced myself that unless there's some additional constraint that I'm missing, or unless I'm not understanding the problem in some other way, that there is not a unique solution to this problem.

[Edit: an example of an "additional constraint" might be something like [itex] V_1 [/itex] and [itex] V_2 [/itex] sharing a common ground, or anything of that nature. But the way it is now (as I presently understand the problem), [itex] V_a [/itex] is unconstrained.]

[Second edit: see my next post. I think there is a way out of this.]
 

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  • #6
collinsmark
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I think there is a way out of this. I think it comes down to the part of the problem statement that says, "First, none of the plates was charged, but then they have been connected to sources of current, ..."

I think from that, we can take that initially all plates were equipotential. And after that point on we can infer that the only variations were changing charge distributions differentially on plates of a given capacitor, but not between the two capacitors (i.e., not overall from one capacitor to another). That should be enough to solve this problem.
 
  • #7
haruspex
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we can take that initially all plates were equipotential.
Yes, I think it is safe to assume that when interleaved the first and third plates have equal and opposite charge, likewise second and fourth.
I got a solution on that basis, but I had a lot of variables: a self-potential at each plate (these all cancel out) and an actual potential at each plate, as well as the two independent per plate charges.
 
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  • #8
I'm not totally sure I understand the configuration. But the way I presently understand it, the physical configuration is as in the figure below,

View attachment 238774

where [itex] V_a [/itex] is the electric potential that we are looking for. But I've pretty much convinced myself that unless there's some additional constraint that I'm missing, or unless I'm not understanding the problem in some other way, that there is not a unique solution to this problem.

[Edit: an example of an "additional constraint" might be something like [itex] V_1 [/itex] and [itex] V_2 [/itex] sharing a common ground, or anything of that nature. But the way it is now (as I presently understand the problem), [itex] V_a [/itex] is unconstrained.]

[Second edit: see my next post. I think there is a way out of this.]
Yes, there is nothing more stated in the problem. I think the author of the problem meant that initially, the switch was open. Honestly, I don't understand what do you mean by
the only variations were changing charge distributions differentially on plates of a given capacitor, but not between the two capacitors
Aren't the capacitor plates supposed to induce the same charges on each plate? Is the field between the plates homogenous? Because if it should be, then the surface charge distributions have to be the same on each plate.(in order the field lines to be equidistant)
 
  • #9
haruspex
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Aren't the capacitor plates supposed to induce the same charges on each plate?
Yes, you can assume the two plates of a given capacitor have equal and opposite charge.
the surface charge distributions have to be the same on each plate.(in order the field lines to be equidistant)
The same for the two plates of a given capacitor, yes, but the two capacitors can have different distributions. The field lines can continue through one plate to the next if necessary.

Did you try the approach I outlined in post #7?
 
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  • #10
Yes, you can assume the two plates of a given capacitor have equal and opposite charge.

The same for the two plates of a given capacitor, yes, but tge two capacitors can have different distributions. The field lines can continue through one plate to the next if necessary.

Did you try the approach I outlined in post #7?
Yes, and I got the right result. I will write it a little bit later, but I used the superposition of the electric fields and the facts you outlined. Thank you for your help
 
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  • #11
SammyS
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Yes, and I got the right result. I will write it a little bit later, but I used the superposition of the electric fields and the facts you outlined. Thank you for your help
Superposition of electric field looks promising to me.
 
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  • #12
ehild
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The inner plates of the two capacitors A(red) an B(blue) can be split, and then we have a chain of three capacitors, C1, C2, C3, with charges q1, q2, q3. The two plates of each original capacitors have equal and opposite charges, as @haruspex suggested, (QA and -QA, QB and -QB) which are distributed between the virtual plates of the new capacitors. See figure.
upload_2019-2-15_16-57-46.png


We can write up the loop equations and also the relation for the charges:

U1-q1/C1-q3/C3=0
U2-q3/C3-q2/C2=0 (Edited)
q3-q2-q1=0
The system of equation is easy to solve.
 

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  • #13
kuruman
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Yes, you can assume the two plates of a given capacitor have equal and opposite charge.
Isn't that a result of charge conservation? Since the capacitors are connected each to its own voltage source and are initially uncharged, whatever number of electrons accumulates on one plate has to come from the other.
 
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  • #14
Delta2
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Isn't that a result of charge conservation? Since the capacitors are connected each to its own voltage source and are initially uncharged, whatever number of electrons accumulates on one plate has to come from the other.
I agree but we also have to state the assumption that the sources cannot store charge inside them.

@ehild, do you care to give a brief explanation of why the two circuits you display are equivalent? I don't understand your intuition behind it, how come you thought of introducing a third new capacitor?
 
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  • #15
I solved the problem, it turned out to be a relatively simple one, yet challenging from the conceptual point of view.
I think that the core idea required for solving this problem lies in the fact that, on the one hand, the electric field between the plates depends on the charges of the plates. Secondary, the field between the plates determines the potential difference. These conditions make up a system of equations.
Let the charge accumulated on the very last plate be ##Q_1##, and on the previous plate ##Q_2##. Exploiting the restraint that the plates were initially uncharged, and the voltage sources only redistribute the charge, on the next 2 plates(I am counting from the last plate to the first one) there will be ##-Q_1## and ##-Q_2##. Then I make use of the superposition principle to calculate the electric field between the plates, in three hollow regions.
Therefore, for the projections of the electric field intensity between the plates I wrote
## E_1= \frac {Q_1} {2S\varepsilon_0}+\frac {Q_2} {2S\varepsilon_0}-\frac {Q_1} {2S\varepsilon_0}+\frac {Q_2} {2S\varepsilon_0}=\frac {Q_2} {S\varepsilon_0}##
##E_2=\frac {Q_1} {2S\varepsilon_0}+\frac {Q_2} {2S\varepsilon_0}+\frac {Q_1} {2S\varepsilon_0}+\frac {Q_2} {2S\varepsilon_0}=\frac {Q_1+Q_2} {S\varepsilon_0}##
##E_3=\frac {Q_1} {2S\varepsilon_0}-\frac {Q_2} {2S\varepsilon_0}+\frac {Q_1} {2S\varepsilon_0}+\frac {Q_2} {2S\varepsilon_0}=\frac {Q_1} {S\varepsilon_0}##
Expressing the potential difference
##V_2=E_3(d-a)+E_2 a##
##V_1=E_1(d-a)+E_2 a##
##V=E_2 a##
Solving this system of equations I got
##V=(V_1+V_2)\frac {a} {d+a}##
 
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  • #16
ehild
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@ehild, do you care to give a brief explanation of why the two circuits you display are equivalent? I don't understand your intuition behind it, how come you thought of introducing a third new capacitor?
You have four parallel plates of the same area and certain distances between each pair of plates. Is not it three capacitors connected? It will be the same three capacitors if you split each plate and connect the parts with a short wire: The pairs will be equipotential with the original plate.
upload_2019-2-15_19-13-3.png
 

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  • #17
ehild
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Solving this system of equations I got
##V=(V_1+V_2)\frac {a} {d+a}##
Very nice solution!
 
  • #19
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In keeping with the original poster's avatar and screen name, the latter circuit diagrams are reminiscent of historical (pre-20th) electrical storage systems. From wikipedia:

"Battery, 18th and 19th century term for a number of capacitors or Leyden jars ..."
"Its invention was a discovery made independently by German cleric Ewald Georg von Kleist..."
 
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  • #20
In keeping with the original poster's avatar and screen name, the latter circuit diagrams are reminiscent of historical (pre-20th) electrical storage systems. From wikipedia:

"Battery, 18th and 19th century term for a number of capacitors or Leyden jars ..."
"Its invention was a discovery made independently by German cleric Ewald Georg von Kleist..."
Ironic coincidence, indeed.
 
  • #21
Delta2
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Worked out the math in @ehild solution and it turns out it gives the same answer as post #15. I used ##C_1=C_2=\epsilon_0\frac{S}{a}## and ##C_3=\epsilon_0\frac{S}{d-a}##. Then I solved not for each charge ##q_i## but for the voltage ##V_3=\frac{q_3}{C_3}## which I believe is the requested voltage.

Interesting that the math work out but to be honest I still cant understand clearly why the two circuits are equivalent...
 
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  • #22
ehild
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Worked out the math in @ehild solution and it turns out it gives the same answer as post #15. I used ##C_1=C_2=\epsilon_0\frac{S}{a}## and ##C_3=\epsilon_0\frac{S}{d-a}##. Then I solved not for each charge ##q_i## but for the voltage ##V_3=\frac{q_3}{C_3}## which I believe is the requested voltage..
Yes, it is correct solution.
Interesting that the math work out but to be honest I still cant understand clearly why the two circuits are equivalent...
An equipotential plane can be replaced with a thin conducting plate. Then surface charge builds up on both surfaces, according to the electric field.
upload_2019-2-16_7-24-49.png


In case of the capacitors in the problem, each plate have some surface charge on both sides. Splitting the plate, the new plates have charge according to these surface charges.
The method is also applied with such problems "fill a parallel plate capacitor half to the distance between the plates with some dielectric. What is the new capacitance? " Here we replace the capacitor with two connected ones, one with dielectric, then other without it.
upload_2019-2-16_7-35-46.png
 

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