# Generation of a Known Charge

Discussion in 'Physics' started by TicTac1231, Apr 26, 2013.

1. ### TicTac1231 Thread Starter New Member

Apr 26, 2013
13
0
Hi all,

Does anyone have an idea how I can go about this? Or maybe ways of quantifying/measuring the charge?

Thanks,

Victor

Last edited: May 31, 2013
2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
If what you wish to do is add a specific amount of static electricity to a ball, that is next to impossible.

If you wish to place a specific amout of charge somewhere else there are two ways that spring to mind.

Firstly if you pass a constant current into some circuit capable of holding charge (eg a capacitor) you can inject a specific charge = current x time.

Alternatively you could try an electrochemical method (electrolysis/deposition) and use a gravimetric chemical analysis to determine the charge passed and then stop at the desired amount.

You need to describe the circumstances further for more help.

Oh and welcome to all about circuits.

3. ### davebee Well-Known Member

Oct 22, 2008
539
46
Charge is voltage times capacitance Q=CV.

The ball will have a fairly fixed capacitance, as long as all nearby objects are always in the same positions, so if you touch the ball to a point at a known voltage, it should always pick up just about the same charge.

Studiot, why do you think that would be so unreliable?

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Hello Davebee,

Firstly what is the ball made of?
An insulating material eg pith or a conducting material, eg copper?

Secondly consider the charge held in computer memory. This has to be 'refreshed' periodically because it leaks away. Leakage is a real enemy of determinism.

So I repeat we need to know more about the circumstances.

Last edited: Apr 26, 2013
5. ### TicTac1231 Thread Starter New Member

Apr 26, 2013
13
0
Hello guys, thanks for the replys and welcome.

Last edited: May 31, 2013
6. ### TicTac1231 Thread Starter New Member

Apr 26, 2013
13
0
Studiot, I also need that charged object to pass to travel in a straight line at a constant velocity, all this is to simulate the induced signal as we would expect from a point charge moving in space. That is why I mentioned a ball becuase there are many ways to get that moving straight.

7. ### TicTac1231 Thread Starter New Member

Apr 26, 2013
13
0
Davebee, so if I have a voltage source, like a 9V battery, I can charge the ball by touching one of the terminals to it?

8. ### davebee Well-Known Member

Oct 22, 2008
539
46
Touching a battery terminal to a ball would put some charge on it, but the amount of charge would be so small that it probably could not be measured and wouldn't affect anything, and there are so many unknowns that the amount of charge would be impossible to predict.

9. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Are you quite sure?

Why would it?

Touching a copper coated steel ball to both terminals simultaneously would certainly put some charge into the ball, but I would strongly recommend against it as this could lead to a dangerous discharge of the battery.

10. ### davebee Well-Known Member

Oct 22, 2008
539
46
Why some charge will transfer - a battery maintains its terminals in a state of unbalanced charge, with electric fields between the two terminals. A conductive ball brought into the field near one of the charged terminals will develop a dipole charge distribution, and when it touches one of the terminals, some of the induced dipole charge will transfer into the battery terminal. When the ball is pulled away, it will have a different overall charge than before.

This behavior is obvious when a ball touches the sphere of a Van de Graaf generator. With a 9 volt battery in free air, the effect will be less, but the principle will be the same; some charge will still transfer. It doesn't matter that neither battery terminal is grounded; there will still be electric fields between the terminals.

That's why there will be "some" charge transfer, to answer your question, studiot.

I get the feeling that there are two conflicting points of view here. One is the engineering point of view, where there is generally considered to be no current unless it affects the operation of a circuit. The other is the physics point of view, where unbalanced charges and electric fields are part of every object, whether conducting or not.

TicTac1231's point of view may be partway between these two, in that he was asking a physics question in a physics forum, but seems to be interested in observable results, so this may not the sort of detail that he'd be interested in.

Getting back to his topic, I don't think there would be enough charge transferred to be detectable.

Clay likes this.
11. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Indeed it is the physics I am concerned with, not the engineering. And this is posted in the physics section.

How much charge does it take to charge up a (perfect) conductor and what is the electric field within that conductor, given perfectly insulating surroundings?

12. ### SplitInfinity Member

Mar 3, 2013
369
9
After reading some of the replies to the topics question I would think that the way in which the ball was created and by what material...ie...the manufacturing Process using a specific material will have much to do as far as the specifics of charging the ball.

In the same way that the design of wire used...such as Twisted 2-Strand Copper Wire....Braided Wire...Inner and Outer Core Wire will all cause variations in the nature of it supplying electrical power to whatever needs it...so if the Ball that you are charging has been created in a Fold/Spiral Method as apposed to using a Mold where say if the ball was iron...molten Iron would be poured into the mold and the Iron would have a relatively equally distributed density of material...as the Fold/Spiral method would have Iron at greater density at certain areas of the ball.

Also the Geometry of what you are charging...and in this case a Ball of unknown construction...such Geometry may cause the charge to flow along specific geomeric lines or Density Lines if present. Also what the ball is in contact with will make a difference as in the example of a Lightning Bolt hitting a Car...even if a person inside was touching a metal stearing wheel due to how the cars tires and due to the construction of the metal exterior...that person would most likely not get shocked.

Split Infinity

13. ### davebee Well-Known Member

Oct 22, 2008
539
46
TicTac1234, what will these charged spheres be used for?

14. ### TicTac1231 Thread Starter New Member

Apr 26, 2013
13
0
Thanks again for the responses.

You are right, I am in-between those two, while its use will be for engineering purposes, I would like to understand the physics behind it.

Last edited: May 31, 2013
15. ### TicTac1231 Thread Starter New Member

Apr 26, 2013
13
0
The balls are also 4.5 mm nominal diameter, but I can certainly look for larger diameter balls which would hold on to a larger charge, and for longer is that would help for the battery discharge idea. Also, I am unsure about the manufacturing of the balls.

16. ### davebee Well-Known Member

Oct 22, 2008
539
46
Maybe you could shoot the balls from a gun where the tip of the barrel is charged to 1000 volts or so.

You'd only need minimal current of that voltage, so it shouldn't be a safety hazard: at one ball per second, 10^3 volts times a capacitance on the order of 10^-12 Farads would give you a current of around 10^-9 amps.

Charge loss: there would be loss via air ionization at voltages over 30 kv per centimeter of curvature of a conductor, so with half a centimeter diameter balls, as long as your voltage is under 15 kv that should not be a problem. Humidity would probably be the next greatest loss mechanism to affect this, but if humidity can be kept low, I'd guess that not much charge would be lost by the balls in air.

If you could measure the current being supplied to the barrel then you could have a measure of how much charge each ball is carrying away.

Oct 22, 2008
539
46
18. ### shortbus AAC Fanatic!

Sep 30, 2009
4,098
1,697
Does this have to have 'balls' to carry the charge? How about a motorized/rotating toothed wheel? The teeth of the wheel could be the same width as the balls diameter, and by having a large distance between teeth, it would simulate the balls rolling past the sensor. The wheel would have a brush to transfer you charge to it, and this would make sure the charge was the same for each tooth.

Also this would eliminate the need to "aim" the balls through the sensor.

19. ### TicTac1231 Thread Starter New Member

Apr 26, 2013
13
0
Davebee, would I expect the gun to give me the projectile the same velocity each time I activate it. I would think that there would be many factors that may affect it.

Last edited: May 31, 2013
20. ### TicTac1231 Thread Starter New Member

Apr 26, 2013
13
0
I was barely able to obtain a signal from there.

Last edited: May 31, 2013