Capacitance of a Parallel Plate Capacitor

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In summary: Hence we see that the total heat generated in a circuit with resistance R is given by:In summary, the homework statement states that there exists a potential difference V between the plates of a parallel plate capacitor of capacitance C. Now it is connected to a cell with potential V. What will be the heat generated?
  • #1
Krushnaraj Pandya
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Homework Statement


There exists a potential difference V between the plates of a parallel plate capacitor of capacitance C. Now it is connected to a cell with potential V. What will be the heat generated?

Homework Equations


Work(on battery) - (energy stored in Capacitor) = Heat produced

The Attempt at a Solution


My logic was that since both have potential V, no charges will flow, but the answer is 2C(V)^2 ; can someone please explain? Thank you
 
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  • #2
Does it say HOW it is connected ?
 
  • #3
BvU said:
Does it say HOW it is connected ?
ohh, there was a diagram which I converted to text here and I just noticed negative terminal of battery is connected to positive plate of capacitor and vice-versa. so will all the charge flow into the cell?
 
  • #4
Where else ? So: Yes! ... but that's not the whole story ...
 
  • #5
BvU said:
Where else ? So: Yes! ... but that's not the whole story ...
physics problems are really exciting to uncover this way...it feels like I'm reading a murder mystery when you say "that's not the whole story...". I really like this forum.
anyway, intuitively speaking, both will have same potential at equilibrium, so does it decrease to zero for both? In that case, energy lost by capacitor is 1/2C(V)^2 and (work done is CV?), some part of my logic doesn't make sense
 
  • #6
Krushnaraj Pandya said:
so will all the charge flow into the cell
As I said: yes. At that point, you have an uncharged capacitor, connected to a cell with potential V (*) . What happens ?

(*) Nothing specific is said about the cell, so we must assume it keeps up the voltage V
 
  • #7
BvU said:
As I said: yes. At that point, you have an uncharged capacitor, connected to a cell with potential V (*) . What happens ?

(*) Nothing specific is said about the cell, so we must assume it keeps up the voltage V
this seems counter-intuitive...so all the charge (Q=CV) flows into the cell, The new potential between the plates is zero and the cell's potential is unaffected...but
1) it must have had done some work to maintain that potential, since charge=CV was flown in at potential V- this work should be C(V)^2
2) The energy lost by capacitor is definitely 1/2 C(V)^2
am I wrong in saying one of these two? perhaps my conceptual clarity is lacking if so. I'd be grateful if you could remove it.
 
  • #8
Pity you already had the answer -- ( naively discharging ##{1\over 2} CV^2## plus charging idem sums up to ##CV^2## so you still miss a factor 2 ).

But you are asking the right questions and I agree with 2). Of 1) I'm not so sure (the ##\Delta V## is not constant) Note that the exercise asks for the heat generated, no matter where.

Krushnaraj Pandya said:
The new potential between the plates is zero and the cell's potential is unaffected...but
This cannot last. The cell doesn't rest until the plates also have a potential difference V. This is what I meant with
BvU said:
What happens ?

Everything is revealed on Hyperphysics (don't forget to click "derive expressions" and "But the ... go ?" )

Krushnaraj Pandya said:
physics problems are really exciting to uncover this way
We usually do not have have the right answer at hand, though ... :wink:
 
  • #9
BvU said:
Pity you already had the answer -- ( naively discharging ##{1\over 2} CV^2## plus charging idem sums up to ##CV^2## so you still miss a factor 2 ).

But you are asking the right questions and I agree with 2). Of 1) I'm not so sure (the ##\Delta V## is not constant) Note that the exercise asks for the heat generated, no matter where.

This cannot last. The cell doesn't rest until the plates also have a potential difference V. This is what I meant withEverything is revealed on Hyperphysics (don't forget to click "derive expressions" and "But the ... go ?" )

We usually do not have have the right answer at hand, though ... :wink:
sorry, I went to meditate for a while, give me a little time to understand capacitors in more detail and so respond accordingly then
 
  • #10
There is a voltage change on the capacitor of 2V ( V - (-V) ) so the energy that the battery must supply (ideally) is ½C(2V)2 = 2CV22

WRT heat that should only be the result of resistance in the circuit and not from creating the electric field between the plates (ideally with no dielectric).
 
  • #11
I suggest since the problem talks about heat generated, we may assume that there is total resistance R in the circuit.

The interesting part is that regardless what is the value of R, we can find that the total heat generated in resistance R is independent of R and it is indeed ##2CV^2##. The time period that this heat will be generated depends on R (theoretically is infinite, but practically is 10RC).

To prove this we can either do the math in a RC circuit (calculate current ##I_1## as function of time for the period where the capacitor is discharging through the battery, then calculate ##\int_0^{\infty}I_1^2(t)Rdt##, then calculate current ##I_2## function for the period where the capacitor is charging and again calculate ##\int_0^{\infty} I_2^2(t)Rdt## and then add ) which i believe is the harder way, or we can think in terms of energy and check what happens to energy provided/absorbed during the discharge and charging process.

Thinking in terms of energy:
For the discharging process your post #7 is correct. The energy provided by the battery is ##\int VIdt=V\int Idt=VQ=CV^2##. The energy provided by the capacitor is ##1/2CV^2##. All this energy is lost as heat in the resistance R. So total heat for this phase is ##3/2CV^2##

For the charging process the energy lost in the resistor as heat will be ##1/2CV^2##. How can you prove that? (think again how much energy the battery provides and how this energy is split into heat in the resistor and electric field energy in the capacitor)
 
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  • #12
BvU said:
charging idem
what do you mean by 'idem'?
 
  • #13
BvU said:
Everything is revealed on Hyperphysics (don't forget to click "derive expressions" and "But the ... go ?" )
I read this and it cleared up a lot of things, thank you
 
  • #14
BvU said:
This cannot last. The cell doesn't rest until the plates also have a potential difference V. This is what I meant with
so, the capacitor discharges but since the cell still has a potential V, the plate that had positive charge initially will start getting negative charge and vice versa till the capacitor is again at a potential V? Am I right?
 
  • #15
gleem said:
There is a voltage change on the capacitor of 2V ( V - (-V) ) so the energy that the battery must supply (ideally) is ½C(2V)2 = 2CV22

WRT heat that should only be the result of resistance in the circuit and not from creating the electric field between the plates (ideally with no dielectric).
Usually the resistance takes up half the energy supplied in the form of heat, I understood your point that the battery must ideally supply 2C(V)^2...but wouldn't that mean that all the energy supplied transformed to heat, since the answer is also 2C(V)^2. where am I going wrong?
 
  • #16
Delta² said:
I suggest since the problem talks about heat generated, we may assume that there is total resistance R in the circuit.

The interesting part is that regardless what is the value of R, we can find that the total heat generated in resistance R is independent of R and it is indeed ##2CV^2##. The time period that this heat will be generated depends on R (theoretically is infinite, but practically is 10RC).

To prove this we can either do the math in a RC circuit (calculate current ##I_1## as function of time for the period where the capacitor is discharging through the battery, then calculate ##\int_0^{\infty}I_1^2(t)Rdt##, then calculate current ##I_2## function for the period where the capacitor is charging and again calculate ##\int_0^{\infty} I_2^2(t)Rdt## and then add ) which i believe is the harder way, or we can think in terms of energy and check what happens to energy provided/absorbed during the discharge and charging process.

Thinking in terms of energy:
For the discharging process your post #7 is correct. The energy provided by the battery is ##\int VIdt=V\int Idt=VQ=CV^2##. The energy provided by the capacitor is ##1/2CV^2##. All this energy is lost as heat in the resistance R. So total heat for this phase is ##3/2CV^2##

For the charging process the energy lost in the resistor as heat will be ##1/2CV^2##. How can you prove that? (think again how much energy the battery provides and how this energy is split into heat in the resistor and electric field energy in the capacitor)
Everything above "thinking...energy" was above me; probably because current electricity is still the next chapter in my textbook, I know that heat generated is independent of R though. Regardless, I should be able to derive the answer from what I have learned.
Anyway, you mention that I'm correct for the discharging process and C(V^2) is supplied by battery and half of that by capacitor; how did you say all this energy is lost as heat though? doesn't a part of 3/2 C(V^2) get used to charge the capacitor again and the other part as heat
Delta² said:
For the charging process the energy lost in the resistor as heat will be 1/2CV^2. How can you prove that? (think again how much energy the battery provides and how this energy is split into heat in the resistor and electric field energy in the capacitor)

I got this part. The battery provides C(V^2), the capacitor takes up half of it and half is lost as heat
 
  • #17
Krushnaraj Pandya said:
Usually the resistance takes up half the energy supplied in the form of heat,

Could you supply the source of this statement?

The energy supplied in charging a battery goes into the work needed to "squeeze" additional charges into the conductors that form the capacitor. Any heat generated would be do to the transporting the charges through a conductor with intrinsic electrical resistance. If you do away with resistance then no heat is generated. Bring charges closer together by itself does not generate heat.
 
  • #18
gleem said:
Could you supply the source of this statement?

The energy supplied in charging a battery goes into the work needed to "squeeze" additional charges into the conductors that form the capacitor. Any heat generated would be do to the transporting the charges through a conductor with intrinsic electrical resistance. If you do away with resistance then no heat is generated. Bring charges closer together by itself does not generate heat.
I read that in charging a capacitor a cell does work CV^2 and energy stored by capacitor is half of that. therefore half is lost as heat. That was the source of my statement. I understood that heat generated is due to the resistance though and not by bringing charges together. But I'm not sure how it'll affect things here any differently mathematically
 
  • #19
Krushnaraj Pandya said:
Anyway, you mention that I'm correct for the discharging process and C(V^2) is supplied by battery and half of that by capacitor; how did you say all this energy is lost as heat though? doesn't a part of 3/2 C(V^2) get used to charge the capacitor again and the other part as heat
where could all this ##3/2CV^2## energy go? The capacitor discharges during the discharging period so it gives away energy, a battery can only provide energy (the battery can't absorb and store energy ) so we got no choice, the energy provided is lost as heat on the resistance R. To see this clearly you have to do the math on the RC circuit as I said in the previous post.
 
  • #20
Krushnaraj Pandya said:
read that in charging a capacitor a cell does work CV^2 and energy stored by capacitor is half of that. therefore half is lost as heat. That was the source of my statement. I understood that heat generated is due to the resistance though and not by bringing charges together. But I'm not sure how it'll affect things here any differently mathematically

In reality heat is generated but the amount is also dependent on the specific characteristics of the cell used. the energy to charge a capacitor is ½CV2 not because of heat but because as the capacitor charges it requires increasingly more work to add charges. The first charge requires basically no work while the last charge must be brought in against the electric field of the all the previous charges.
 
  • #21
Delta² said:
where could all this ##3/2CV^2## energy go? The capacitor discharges during the discharging period so it gives away energy, a battery can only provide energy (the battery can't absorb and store energy ) so we got no choice, the energy provided is lost as heat on the resistance R. To see this clearly you have to do the math on the RC circuit as I said in the previous post.
Here is my step-wise understanding of what is happening-
1) capacitor discharges- 1/2 CV^2 lost as heat
2)Battery does work CV^2 to charge capacitor fully again
3)of this work, half is stored up by capacitor and half is lost as heat
At the end then we have a charged capacitor and CV^2 of lost heat. What step did I miss or which one is wrong?
As for the RC circuits...I'm afraid that's in the next chapter so I don't have even basic knowledge regarding them yet. But I'm sure there must be a way to understand this by what I already know since it is given in the book before RC circuits are introduced.
 
  • #22
gleem said:
In reality heat is generated but the amount is also dependent on the specific characteristics of the cell used. the energy to charge a capacitor is ½CV2 not because of heat but because as the capacitor charges it requires increasingly more work to add charges. The first charge requires basically no work while the last charge must be brought in against the electric field of the all the previous charges.
Ah! right, I understand the integral used in the derivation now. Thank you for clearing up this concept in my mind.
 
  • #23
Krushnaraj Pandya said:
Here is my step-wise understanding of what is happening-
1) capacitor discharges- 1/2 CV^2 lost as heat
2)Battery does work CV^2 to charge capacitor fully again
3)of this work, half is stored up by capacitor and half is lost as heat
At the end then we have a charged capacitor and CV^2 of lost heat. What step did I miss or which one is wrong?
As for the RC circuits...I'm afraid that's in the next chapter so I don't have even basic knowledge regarding them yet. But I'm sure there must be a way to understand this by what I already know since it is given in the book before RC circuits are introduced.
step 1) is not completely described. a correct 1) would be :
1) Capacitor discharges, so 1/2 CV^2 lost as heat, but also charge Q=CV has flown through the battery V, so battery has done work W=QV=CV^2 on this charge and this work is eventually lost again as heat in the resistance, so total CV^2+1/2CV^2 lost as heat.

I did a mistake earlier when I said that a battery can't absorb energy, a battery is considered to absorb energy when the current through battery is from the positive to the negative terminal. But if you do a scheme and connect the battery and the capacitor with reversed polarities you ll see that the current through battery during the discharging process, is from the negative to the positive terminal, so the battery gives away energy.
 
  • #24
Delta² said:
step 1) is not completely described. a correct 1) would be :
1) Capacitor discharges, so 1/2 CV^2 lost as heat, but also charge Q=CV has flown through the battery V, so battery has done work W=QV=CV^2 on this charge and this work is eventually lost again as heat in the resistance, so total CV^2+1/2CV^2 lost as heat.

I did a mistake earlier when I said that a battery can't absorb energy, a battery is considered to absorb energy when the current through battery is from the positive to the negative terminal. But if you do a scheme and connect the battery and the capacitor with reversed polarities you ll see that the current through battery during the discharging process, is from the negative to the positive terminal, so the battery gives away energy.
Ah! I understand clearly now. Thank you very much for your patience in explaining this to me.
Also, I meant to ask- are all threads permanently archived on the site? If someday I wish to revisit this question, it'd be nice just to be able to open it again a year later and understand things easily
 
  • #25
Krushnaraj Pandya said:
Ah! I understand clearly now. Thank you very much for your patience in explaining this to me.
Also, I meant to ask- are all threads permanently archived on the site? If someday I wish to revisit this question, it'd be nice just to be able to open it again a year later and understand things easily

Well as far as I know yes. Well to be accurate almost all threads, there are very few threads that could be deleted for various reasons (like if they become too religious, or attempt to discuss non mainstream physics , or because they are pure nonsense).
 
  • #26
Delta² said:
Well as far as I know yes. Well to be accurate almost all threads, there are very few threads that could be deleted for various reasons (like if they become too religious, or attempt to discuss non mainstream physics , or because they are pure nonsense).
I wanted to ask a few things, is it alright if I send a message to your inbox?
 
  • #27
Yes sure go ahead.
 
  • #28
A little simpler way of getting the answer is to look at just the initial and final states of the system rather than looking at the discharging and charging processes separately. The energy stored in capacitor is unchanged since the sign of the potential difference across the capacitor ##\Delta V_\text{C} = \pm V## doesn't matter, so any work done by the battery has to be dissipated as heat. The battery moves a charge 2Q through a potential ##V##. How much work does it do?
 
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  • #29
vela said:
A little simpler way of getting the answer is to look at just the initial and final states of the system rather than looking at the discharging and charging processes separately. The energy stored in capacitor is unchanged since the sign of the potential difference across the capacitor ##\Delta V_\text{C} = \pm V## doesn't matter, so any work done by the battery has to be dissipated as heat. The battery moves a charge 2Q through a potential ##V##. How much work does it do?
That really is much much simpler, thank you for your insight...
 

Related to Capacitance of a Parallel Plate Capacitor

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is measured in units of Farads (F).

2. How do you calculate capacitance?

Capacitance is calculated by dividing the amount of electric charge (in Coulombs) by the potential difference (in Volts) between the two conductors. This can be represented by the equation C=Q/V, where Q is the charge and V is the potential difference.

3. What is a simple capacitance problem?

A simple capacitance problem involves finding the capacitance of a system with known values of charge and potential difference, or finding one of these values given the other and the capacitance. It may also involve finding the energy stored in the system.

4. What is the relationship between capacitance and distance?

Capacitance is inversely proportional to the distance between the two conductors. This means that as the distance increases, the capacitance decreases, and vice versa.

5. How does the material between the conductors affect capacitance?

The material between the conductors, known as the dielectric, affects capacitance by either increasing or decreasing it. Materials with high permittivity, such as glass or plastic, increase capacitance, while materials with low permittivity, such as air or vacuum, decrease capacitance.

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