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Better understanding of capacitor charging

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Kajunbee

What causes one plate of capacitor to not be able to accept more charge. I had assumed it was because of the force needed to push charge off the other plate. But after I thought about it this does not seem to be the case. Even as charge on one plate is forcing charge off the other plate. There is also and attraction to the positive charges that are remaining. And analogy I guess would be a air compressor and the suction and discharge are both connected to the same tank. If this is truly the case where does the force that resist one plate excepting more charge come from. Hopefully someone can clear up my misconceptions.

FvM

Super Moderator
Staff member
Not clear which effect you are asking about. Presuming voltage independent capacitance, which is roughly achieved by most capacitors, stored charge is proportional to capacitor voltage,

Q = C*U

In which situation are you seeing a capacitor not being "able to accept more charge"? As long as the dielectric doesn't suffer electrical breakdown, you can send more charge to the capacitor by increasing the voltage.

Kajunbee

Kajunbee

Points: 2

Ratch

What causes one plate of capacitor to not be able to accept more charge. I had assumed it was because of the force needed to push charge off the other plate. But after I thought about it this does not seem to be the case. Even as charge on one plate is forcing charge off the other plate. There is also and attraction to the positive charges that are remaining. And analogy I guess would be a air compressor and the suction and discharge are both connected to the same tank. If this is truly the case where does the force that resist one plate excepting more charge come from. Hopefully someone can clear up my misconceptions.

Before you can gain a better understanding of how a capacitor works, you must first gain a basic knowledge of it. The use of correct terminology is essential to avoid confusion. A capacitor contains the same net charge whether there is zero volts or 1000 volts across it. That is because for every coulomb of charge added to one plate by the negative voltage, the same charge is subtracted from the opposite plate by the positive voltage. It takes energy to imbalance the charge on the plates, and this energy is stored in the electrostatic field between the two charge differences where the dielectric resides. You should not use the unscientific slang description by saying a capacitor is "charged" or "discharged". The net charge does not change. You should instead refer to it being energized or de-energized. Mobile positive charges do not exist because holes do not exist in conductive metals, and the metallic core ions that lose their electrons are not mobile, and remain fixed within their metallic structure. So when you crank up the voltage, and do not exceed the breakdown of the dielectric, you cause a greater separation of charges, but the net charge remains the same as it was at zero volts. However, the energy stored in the capacitor increases. It becomes more energized, not charged. Ask if you have any questions about this.

Ratch

Last edited:
Kajunbee

Kajunbee

Points: 2

Aussie Susan

Or are you talking about the relationship between capacitance, change of voltage and instantaneous current:

i = C * dv/dt

If by "accepting charge" you mean the current flow (of electrons that carry the charge) then the formula shows how that varies with voltage and time.
As that involves a bit of calculus I can see why it might be a little harder to understand what is happening.
Susan

Kajunbee

Kajunbee

Points: 2

c_mitra

What causes one plate of capacitor to not be able to accept more charge.

Consider a sphere and add some charge to it. The charge may be considered concentrated at the centre and will have a field at the surface and a corresponding potential.

Now add some more charge; it has to be moved against the existing field and the potential.

If you want to focus on a two plate capacitor, just consider one end being grounded.

Kajunbee

Kajunbee

Points: 2

Ratch

Or are you talking about the relationship between capacitance, change of voltage and instantaneous current:

i = C * dv/dt

If by "accepting charge" you mean the current flow (of electrons that carry the charge) then the formula shows how that varies with voltage and time.
As that involves a bit of calculus I can see why it might be a little harder to understand what is happening.
Susan

"Current flow" is another one of those unscientific slang terms. It literally means "charge flow flow", which is redundant and ridiculous. One should just say current exists or current is present or just plain charge flow or just plain current. Charge does not flow twice.

If you hook up an ammeter to each lead of a capacitor, and apply a voltage to energize the capacitor, you will observe both ammeters showing a current in each of the two leads. However, no current exists through the capacitor. The charge is accumulating on one plate and depleting on the opposite plate, but no charge is passing through the capacitor. Obviously this cannot go on indefinitely. Either the charge flow will decrease as the back voltage builds up, or more voltage will have to be applied to keep up the charge gain/loss rate.

Ratch

Kajunbee

Kajunbee

Points: 2

Aussie Susan

Hopelessly Pedantic
OK, so I got the '(' in the wrong place - "...you mean the current (flow of electrons that..." - Better
Susan

Kajunbee

The information given has been extremely helpful. I have a much better understanding now but there are couple things I'm unsure about. We start with a "energized " capacitor and the plates are a small distance apart. Let's say the voltage is 10 volts and the terminals are not connected. We now decrease the distance between the plates. If I am correct the voltage should drop. I understand that since the plate is closer the attraction between the plates should be stronger. The same way a north and south facing magnet have a stronger attraction the closer they get. Is the voltage drop associated with the fact that the positive charges are immobile.

c_mitra

Let's say the voltage is 10 volts and the terminals are not connected. We now decrease the distance between the plates. If I am correct the voltage should drop.

Always use some model to get insight.

Consider the same charge applied to two spheres with different radii; in both cases, the total charge can be considered to be present at the centre.

Electric field and the potential will be different because they have different radii; the larger sphere will have a lower voltage.

In the Wimshurst machine (high voltage generator), the distance between the plates is changed to produce high voltages.

Kajunbee

Kajunbee

Points: 2

Ratch

OK, so I got the '(' in the wrong place - "...you mean the current (flow of electrons that..." - Better
Susan

Yes, current is the movement rate of charge carriers (electrons in this case).

Ratch

- - - Updated - - -

The information given has been extremely helpful. I have a much better understanding now but there are couple things I'm unsure about. We start with a "energized " capacitor and the plates are a small distance apart. Let's say the voltage is 10 volts and the terminals are not connected. We now decrease the distance between the plates. If I am correct the voltage should drop. I understand that since the plate is closer the attraction between the plates should be stronger. The same way a north and south facing magnet have a stronger attraction the closer they get. Is the voltage drop associated with the fact that the positive charges are immobile.

I don't think you understand what voltage is. It takes energy to gather charge carriers together in one place. It takes even more energy if there is an increase of charge to gather, and even more energy if the charge carriers are packed closer together. Voltage is a measure of the energy density of the charge, and its unit in the MKS system is joules/coulomb. When one end of a conductor has a higher voltage (energy density per unit of charge) it will try to equalize with the other end of the conductor by charge movement from the higher voltage to the lower voltage. That will cause a current to exist as long as there is a conduction path and a voltage difference.

In the case of the capacitor you describe, moving the plates closer will cause the capacitance to increase. Increasing the capacitance will cause the voltage to decrease if the same charge imbalance is maintained (V = Q/C). The attraction of the plates have nothing to do with the voltage across the capacitor or the mobility of the charge carriers. I would think the attraction is minuscule anyway, perhaps attracting some dust particles. Don't confuse force with voltage or energy. Voltage is not force. There can be a voltage difference across a capacitor, but a voltage drop at one lead is a voltage rise referenced from the other lead. Therefore, you have to establish a reference before you can say the voltage is rising or dropping.

Ratch

Kajunbee

Kajunbee

Points: 2

c_mitra

Voltage is a measure of the energy density of the charge

This is wrong.

Charge is associated with a field (electric field in this instance)- this is a force field that exerts force on another charge placed in the field. The force is given by the Coulmb's law.

Force is a vector and the field is a vector field.

Every point in this electric field is associated with a potential such that the derivative of the potential is the electric field. Equivalently, this is also the work done on a unit test charge to be moved from the test point to infinity (electric field is zero only at infinity).

For a static charge, the field is static and the potential is a scalar. If the charge is moving, the potential becomes also a vector.

Kajunbee

Kajunbee

Points: 2

Ratch

This is wrong.

I should have said that "voltage is the energy density per unit charge". That is my corrected statement. That is why voltage units are joules/coulomb.

Charge is associated with a field (electric field in this instance)- this is a force field that exerts force on another charge placed in the field. The force is given by the Coulmb's law.

What you say above is basically true, but it does not contradict what I aver. An electric field is a region of influence to electrical charges. An electric field can only be detected by the force it exerts on a electrical charge, but its influence alone cannot be said to be a force.

Force is a vector and the field is a vector field.

Yes, that is basic physics.

Every point in this electric field is associated with a potential such that the derivative of the potential is the electric field. Equivalently, this is also the work done on a unit test charge to be moved from the test point to infinity (electric field is zero only at infinity).

What you said directly above is also basically true. I might add that potential needs a reference point that is usually taken to be an infinite distance away, but not always. The above still does not contradict what I said about voltage.

For a static charge, the field is static and the potential is a scalar. If the charge is moving, the potential becomes also a vector.

Voltage is always a scalar because it has no direction. If one moves a charge in a circle around a non-infinite reference point, the voltage will be constant because it depends only on the distance from the reference point and not the direction around the circle. Voltage is a measure of the energy density per unit charge, and density has no direction. Now, you can say that the voltage changes when the the charge is moved in a certain direction, but voltage change is related to electric field change, not the voltage itself which is always scalar.

Ratch

c_mitra

I should have said that "voltage is the energy density per unit charge". That is my corrected statement. That is why voltage units are joules/coulomb.

Why not joules/liter/coulomb?

Basically you are offering a circular argument.

Energy is basically defined in terms of force (work done = energy = force X distance)

Force is the causative agent of motion (Newton)

Energy is defined as either kinetic or potential.

Potential energy is always due to a force field.

Potential is the quantity whose derivative is the force (Lagrange)

Defining potential in terms of energy is therefore a circular concept.

Kajunbee

Kajunbee

Points: 2

Ratch

Why not joules/liter/coulomb?

Because voltage is not a volumetric density. It is an energy density, similar to temperature (heat energy/mass)

Basically you are offering a circular argument.

How so?

Energy is basically defined in terms of force (work done = energy = force X distance)

Mechanical energy is, but there are many different forms of energy. Heat, radiation, electrical, spring tension, or anything with the ability to do work.

Force is the causative agent of motion (Newton)

Yes, for mechanical action.

Energy is defined as either kinetic or potential.

A promise to do work and really doing something.

Potential energy is always due to a force field.

Nope, a gallon of gasoline has a lot of potential energy, but no force field.

Potential is the quantity whose derivative is the force (Lagrange)

The derivative with respect to what variable?

Defining potential in terms of energy is therefore a circular concept.

How so?

The concept of energy density should not be difficult to understand. For instance, hydroelectric power (dams) is known for being a high energy density source. Wind power (windmills) is knows as a thin energy source. When charge carriers like electrons are gathered together, they want to go somewhere else. It takes energy to force them together, and the energy required per unit charge is the voltage (joules/coulomb).

Ratch

Kajunbee

Kajunbee

Points: 2

c_mitra

It is an energy density, similar to temperature (heat energy/mass)

Your concept of temperature is flawed; it is similar to a potential that determines the direction of flow of heat from one substance to another.

And temp cannot be expressed in units of the three basic ones: mass, length and time.

Temp is an important abstract concept of physics. Start with the ideal gas laws.

Ratch

Your concept of temperature is flawed; it is similar to a potential that determines the direction of flow of heat from one substance to another.

And temp cannot be expressed in units of the three basic ones: mass, length and time.

Temp is an important abstract concept of physics. Start with the ideal gas laws.

There is more to physics than MLT. There is a bunch of energy forms, fields of all kinds, force, torque, momentum, nuclear, and who knows what else.

Temperature is a heat energy density of a mass. The more heat energy you pump into a defined mass, the higher the temperature. Increase the mass without adding more energy and the temperature drops. Heat energy flows from the higher energy density to the lower energy density through a non-insulated path. Seems rather straight forward to me.

Ratch

Kajunbee

Points: 2

Kajunbee

Kajunbee

Points: 2

Akanimo

The information given has been extremely helpful. I have a much better understanding now but there are couple things I'm unsure about. We start with a "energized " capacitor and the plates are a small distance apart. Let's say the voltage is 10 volts and the terminals are not connected. We now decrease the distance between the plates. If I am correct the voltage should drop. I understand that since the plate is closer the attraction between the plates should be stronger. The same way a north and south facing magnet have a stronger attraction the closer they get. Is the voltage drop associated with the fact that the positive charges are immobile.

The voltage will not drop. Instead, it will go up.

When you decrease the distance between the plates, the field becomes stronger. This means that the capacitance decreases because you now need less energy to keep the charges on the plates as there's already a stronger attraction between the unlike charges on the two plates.
This also implies that the voltage increases since the charges are still intact. This could be easily seen with this known formula:

Q=CV

Since Q is constant, if C decreases, V must increase proportionately. A lower capacitance requires a lesser amount of charge to increase the voltage by 1V. Remember?

FvM

Super Moderator
Staff member
Kajunbee talks about decreasing capacitor plate distance. Capacitance increases and voltage drops respectively.

Kajunbee

Points: 2