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.
What causes one plate of capacitor to not be able to accept more charge.
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
OK, so I got the '(' in the wrong place - "...you mean the current (flow of electrons that..." - BetterHopelessly Pedantic
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.
OK, so I got the '(' in the wrong place - "...you mean the current (flow of electrons that..." - Better
Susan
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.
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.
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.
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.
There is more to physics than MLT.
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.
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