# fundamental behavior of capacitor's plates and the charges in response to sudden changes in voltage?

#### TarikElec

Joined Oct 17, 2019
122
Hello everyone,
I have been trying to understand fully the behaviour of the capacitor and over its two plates when exposed to sudden changes in voltage, and I divide my request for help into two questions:

1- How and why when we apply a sudden voltage over a capacitor C1_Plate_1, the C1_Plate2 follows as:
at the positive edge V1(0V-> 1V), the C1_Plate_2=1V
at the negative edge V1(1V-> 0V), the C1_Plate_2=-1V
I tried to use maxwell equations to understand this behaviour, but I failed
also, I have looked for documentation to explain these phenomena from physics perspective regarding the Electric field and charge ,but no success.

2-When two capacitors in parallel are charged by a continuos source voltage V during the whole operation.
-t=0--->> V0=0
so, Q=C*V, (V=V1=V2),
Qtot=Q1+Q2
Ctot=C1+C2
V1=Q1/C1=V2=Q2/C2=Qtot/Ctot
So finally,
Q1=V*C1/Ctot
Q2=V*C2/Ctot

-at t=t1-->> V0=2
Here are my questions:
does the V0 create more charge, which means Qtot'=Qtot+Q'? If yes, how is it calculated and explained physically?
How to calculate the new voltage V2?

Any reading that will help me get the answer by myself would be appreciated. Because I want to do it logically and arrive at the right final results

#### crutschow

Joined Mar 14, 2008
34,455
How to calculate the new voltage V2?
In the bottom schematic, V2 is determined by the voltage source V, so it stays at 2V.

All the voltages can be determined by using the basic capacitor charge vs. voltage formula of V = Q/C.

#### LowQCab

Joined Nov 6, 2012
4,075
The exact specifics of how a Capacitor works is not demonstrable with a Math-Formula.
That's why there are different "types" of Capacitors,
each with their own "personality", and set of "expected" performance-specifications.
.
.
.

#### WBahn

Joined Mar 31, 2012
30,065
Hello everyone,
I have been trying to understand fully the behaviour of the capacitor and over its two plates when exposed to sudden changes in voltage, and I divide my request for help into two questions:

1- How and why when we apply a sudden voltage over a capacitor C1_Plate_1, the C1_Plate2 follows as:
at the positive edge V1(0V-> 1V), the C1_Plate_2=1V
at the negative edge V1(1V-> 0V), the C1_Plate_2=-1V
I tried to use maxwell equations to understand this behaviour, but I failed
also, I have looked for documentation to explain these phenomena from physics perspective regarding the Electric field and charge ,but no success.
Uh.... it is a direct consequence of the Conservation of Energy.

The energy stored in a capacitor is related to the voltage across it. Change the voltage, and you change the energy. Since energy can neither be created nor destroyed, merely converted from one form to another, that means that the energy has to be a continuous quantity -- it can change quickly, but it cannot change instantaneously. Since capacitor voltage is related to energy, that means that the voltage across a capacitor cannot change instantly.

So if you have a capacitor that has a voltage of 100 V across it and you instantly change the voltage on one plate by 10 V, the voltage of the other place will change by 10 V in the same direction. This is required in order to keep the voltage across the capacitor the same across this infinitesimally small amount of time.

This behavior is crucial to the functioning of many devices, such as a flying capacitor voltage doubledr

Similarly, the energy in an inductor is related to the current through it, so current in a conductor cannot change instantly and must be a continuous quantity. This is used in things like automotive ignition systems and switch-mode power supplies.

#### MisterBill2

Joined Jan 23, 2018
18,538
This post and thread have the appearance of schoolwork. not applicable to real world use.

#### nsaspook

Joined Aug 27, 2009
13,297
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#### WBahn

Joined Mar 31, 2012
30,065
This post and thread have the appearance of schoolwork. not applicable to real world use.
Lot's of circuits use this concept in the read world. I gave one example, namely a flying capacitor voltage doubler. Other examples are charge pumps used to produce multiple supply voltages on an IC from a single supply. Ever use something like a UART chip that is powered from the system's 3.3 V or 5 V supply but yet somehow magically produces -12 V and +12 V output signals? It is also the fundamental concept behind switched-capacitor filters.

#### WBahn

Joined Mar 31, 2012
30,065
if you say so. Not freshman schoolwork, certainly. perhaps even 300 level.
The fundamental explanation behind this behavior is covered in Physics II (Intro E & M) and I certainly saw it in my Circuits I course and my Physics Analog Electronics course, both of which were sophomore year.

The application where I first put it to use was creating a short downward pulse at the falling edge of a logic signal to trigger a 555-based one-shot whose output was shorter than the width of the input signal. This sounds like it might be described as "capacitor coupled digital edge triggers for rising or falling signal edges."

#### nsaspook

Joined Aug 27, 2009
13,297
if you say so. Not freshman schoolwork, certainly. perhaps even 300 level.
Yes, I say so. It's pretty basic physics and electronics. Take a look at that classic book TTL cookbook. It's loaded with plenty of edge generation and detection examples with trigger design examples that use simple RC physics.

Think about energy first in circuits.

#### MisterBill2

Joined Jan 23, 2018
18,538
Lot's of circuits use this concept in the read world. I gave one example, namely a flying capacitor voltage doubler. Other examples are charge pumps used to produce multiple supply voltages on an IC from a single supply. Ever use something like a UART chip that is powered from the system's 3.3 V or 5 V supply but yet somehow magically produces -12 V and +12 V output signals? It is also the fundamental concept behind switched-capacitor filters.
I have used capacitor charge pumps a used the trigger pulse scheme quite a few times, even solved problems for folks in this forum with it. But not in the last three months. And I know which kinds of capacitor to use.

#### TarikElec

Joined Oct 17, 2019
122
In the bottom schematic, V2 is determined by the voltage source V, so it stays at 2V.
I got it wrong on the circuit 2 and you are right the voltage stays V2
All the voltages can be determined by using the basic capacitor charge vs. voltage formula of V = Q/C.
here, the issue is at the plates

#### TarikElec

Joined Oct 17, 2019
122
This post and thread have the appearance of schoolwork. not applicable to real world use.
this is not a homework. I want to understand fully the concept of charge redistribution and I made this circuit as simple as possible to get the concept first

#### WBahn

Joined Mar 31, 2012
30,065
this is not a homework. I want to understand fully the concept of charge redistribution and I made this circuit as simple as possible to get the concept first
In the instant that the voltage at one plate is changed, there is no charge redistribution. All that has happened is the one plate was forced to a different potential. The charge on the plates is still there, which still results in the same voltage difference across the plates, and his whatever voltage change one plate undergoes is echoed at the other plate.

#### TarikElec

Joined Oct 17, 2019
122
Uh.... it is a direct consequence of the Conservation of Energy.

So if you have a capacitor that has a voltage of 100 V across it and you instantly change the voltage on one plate by 10 V, the voltage of the other place will change by 10 V in the same direction. This is required in order to keep the voltage across the capacitor the same across this infinitesimally small amount of time.
Thank you very much.
Summary, energy does not change instantaneously in capacitor because Energy in capacity is stored in the electric field (Energy=1/2*C*(E/d)^2) and to change the electric field we need to bring electric charges into the two plates which is impossible instantaneously and this leads to that the sudden change voltage in one plate bring the same other plate to the same voltage change for final purpose to keep the energy non-changeable instantaneously!!
P.S: same things with inductors but with current(back EMF as opposing the quick change of current)

#### WBahn

Joined Mar 31, 2012
30,065
Thank you very much.
Summary, energy does not change instantaneously in capacitor because Energy in capacity is stored in the electric field (Energy=1/2*C*(E/d)^2) and to change the electric field we need to bring electric charges into the two plates which is impossible instantaneously and this leads to that the sudden change voltage in one plate bring the same other plate to the same voltage change for final purpose to keep the energy non-changeable instantaneously!!
P.S: same things with inductors but with current(back EMF as opposing the quick change of current)
That's about it -- the one thing I would say (and I suspect this is just a translation-to-English issue) is that, as written, you are claiming that because energy is stored in the electric field is the reason that energy does not change instantaneously. But the reason that energy cannot change instantaneously is because of the conservation of energy and an instantaneous change would require infinite power (since power is the rate of change of energy from one form to another).

In the real world, we don't have pure capacitances or pure inductances -- there's always some aspect of both, even if unintentionally. But we can make the one we don't want extremely small to the point that the difference between "instantaneous:" and "really, really, fast" are indistinguishable for nearly all purposes.

#### TarikElec

Joined Oct 17, 2019
122
That's about it -- the one thing I would say (and I suspect this is just a translation-to-English issue) is that, as written, you are claiming that because energy is stored in the electric field is the reason that energy does not change instantaneously. But the reason that energy cannot change instantaneously is because of the conservation of energy and an instantaneous change would require infinite power (since power is the rate of change of energy from one form to another).

In the real world, we don't have pure capacitances or pure inductances -- there's always some aspect of both, even if unintentionally. But we can make the one we don't want extremely small to the point that the difference between "instantaneous:" and "really, really, fast" are indistinguishable for nearly all purposes.
to be honest I did not understand well the connection between the law of conservation of law and the reason cannot change instantaneously.
If I give you a mechanical example of a ball that has kinetic and potential energy. if it falls from A to B which means its potential energy is decreasing while its kinetic energy is increasing. Here is the thing: what if I stopped brutally the ball by an external force that convert the kinetic energy immediately to 0, isn't that an instantaneously change of energy?

#### nsaspook

Joined Aug 27, 2009
13,297
Your instantaneously (stopped brutally) change is actually an “an arbitrarily small period of time" derived by you, not physics. If change happened with zero time there would be no universe as we know it today because there would be no action at a distance forces because there would be no distances (locality) because there would be no causality.

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#### Papabravo

Joined Feb 24, 2006
21,225
If you examine what happens to an object when its momentum is dissipated rapidly, there is elastic deformation and rebound. If there is enough kinetic energy and the elastic limit is reached, then the deformation becomes plastic and or the object explodes.

#### WBahn

Joined Mar 31, 2012
30,065
to be honest I did not understand well the connection between the law of conservation of law and the reason cannot change instantaneously.
If I give you a mechanical example of a ball that has kinetic and potential energy. if it falls from A to B which means its potential energy is decreasing while its kinetic energy is increasing. Here is the thing: what if I stopped brutally the ball by an external force that convert the kinetic energy immediately to 0, isn't that an instantaneously change of energy?
It can't happen. It would require an infinite force to do so, and an infinite force would break whatever what applying the force, whether it was a baseball bat or a slap of concrete.

In reality, what happens is that either or both objects deform (or break) and there is a finite amount of time for the energy to be converted from kinetic energy to some other form or combination of forms, such as heat.