# Voltage and charged parallel plates-Confused!

1. Jan 6, 2010

### SMc21

So, we learned about electricity in Physics a few months ago. One thing from the unit is still bugging me that I don't understand: The Millikan Oil Drop Experiment and the idea of charged parallel plates.

Apparently, you can "apply a voltage" to charged parallel plates to create a field between them. I don't get it, though: Isn't voltage a property of the field, and not the other way around?
Also, by varying the distance between the plates while keeping voltage constant, you can apparently alter the field strength. But isn't field strength dependent on charge? The only thing I can think of is that the amount of charge on each plate is changing, but that raises more questions: Where does the charge come from or go to when distance is altered?

Any help is greatly appreciated.

2. Jan 6, 2010

### Prologue

We use the terminology 'apply a voltage' because the power supply that you use has a constant voltage, say 5V. What does that mean? It means that if you did a line integral from one terminal to the other, you would end up with 5V. That is the definition of voltage. So, the field is always there, it may not be constant, but it is there in some shape or form. When you attach plates to the terminals and arrange them in parallel, the power supply works its *** off to be sure that the line integral between the two is still 5V. It does this by stealing charge from one plate and distributing it on the other. So, yes, the charge is changing. When the plates are far apart, the distance between them is big, therefore the line integral is along a long path, and so the field can be small. Remember that the field is constant along a line between the two plates so the line integral is merely separation*E-Field=5V. Now, when you bring them close the distance is small, so the field has to be big. But, remember that the field strength is ONLY determined by the amount of charge on the plates, so in order to generate the big field it needs to crank up the amount of charge.

All of this 'metering' and checking, is done by the power supply, its sole purpose in life it to make sure that the line integral between two points on the different terminals/plates is 5V. IT does that by moving charge from one spot to another (it has to do work to do this).

3. Jan 6, 2010

### vertices

The voltage is simply the potential difference across the plates. A field is a region of space with a particular value ascribed to it. We can talk about the temperature field of your body - each part of your body has a different temperature.

The electric field strength, E, is simply the amount of force a colomb of charge will feel if you place it in the field (well, the force a test charge, dq, will feel to be more precise, ie we define the electric field E=Force/dq) and it is normally a function of position.

The work done by a charge dq moving through a pd dV is F*dr which equals -dq*dV (where dr is the distance it has moved) so we get that E=F/dq=-dV/dr. For the the two plates, dV/dr is just constant in the region between them (mathematically, the field strength is just Voltage/distance between plates) - that's why you need to increase the distance, keeping V fixed, to change E.

4. Jan 6, 2010

### SMc21

Ah, so the plates function as an "extension" of the battery?
I was picturing the plates as behaving like a "circuit" (The battery uses its potential difference to send charges through them). I guess that in reality, they behave more like the terminals of the battery. Am I correct here?

5. Jan 6, 2010

### Prologue

The plates are at the same voltage as the terminals, this is because metal is an equipotential surface (ideally).

6. Jan 6, 2010

### SMc21

So in this case, the function of the battery is basically to distribute charges in the plates to create a field with voltage equal to its own?

7. Jan 6, 2010

### Prologue

It is more like, 'that is just what happens'. The purpose of the battery is to maintain a voltage, that is it. It achieves its goal by distributing charges.

8. Jan 13, 2010

### SMc21

Alright. One more question, concerning the behaviour of the battery:

When a battery is connected to a current, charges are "pumped" from one terminal to the other inside the battery and gain energy. They then travel through the circuit and lose electric potential energy, and the process repeats.

However, when a battery is connected to parallel plates, the charges can't "travel" since the plates aren't connected. They stay on the plates at the energy level given to them by the battery and thus generate an electric field between them. In this sense, the plates are like extensions of the terminals.

Am I using faulty reasoning here, or am I somewhat on the ball?

9. Jan 14, 2010

### SMc21

Maybe I should clarify why I'm confused. I'm trying to connect the idea of a battery here with the idea of a battery in current electricity:

When I studied current electricity, I learned that a battery's purpose was to "pump" current from one terminal to another. To do this, it basically creates an electric field inside itself. The current is moved in the opposite direction of the field, giving it energy. It then flows through the circuit and loses electric energy. The process repeats itself once the current reaches the oppositely-charged terminal.

In the case of the Millikan Experiment, though, the battery's purpose is to maintain a constant voltage across the plates connected to its terminals. It does this by varying the charge in each plate depending on their distance apart. This is what I meant when I said the plates were like "extensions" of the battery's terminals: The terminals/plates are both given a constant voltage.

So, the battery can either be a "pump" or a voltage regulator depending on the situation? If we connect a wire from one plate to the other, do they still maintain their potential difference (Acting as "terminals"), or do they become part of the current-carrying conductor?