# Creating an electric field

1. Nov 25, 2008

### simo

I want to create an electric field that will cause a copper BB to move. So far I have tried using a 9V battery with the positive end attached to a metal sheet and the negative attached to another sheet, like a capacitor. Unfortunately this hasn't worked.
Any ideas?

2. Nov 25, 2008

### Integral

Staff Emeritus
Before you could move a copper BB in an electric field you would need to induce a charge on it.

3. Nov 25, 2008

### simo

The field should cause a charge separation inside the BB because its a conductor, right?

4. Nov 26, 2008

### m.e.t.a.

You're right, simo, there would be an induced charge separation inside the metal BB because it is a conductor. In this case you would be able to pull the BB around with electric fields, but not push it. I think your problem is that the E-field you are generating is much too small to exert an appreciable force on your copper pellet over any kind of distance. I'll get back to you later with the appropriate calculation if I get the time. But as a rough guess, you're going to need at least a few hundred volts. Either that or you'll need a very slick surface so that small accelerations of the BB can occur without being swallowed by friction.

5. Nov 26, 2008

### HallsofIvy

There would be a charge separation within the BB but since it is rigid, you cannot use that charge separation. The BB will move according to the net force on it which will be 0.

6. Nov 26, 2008

### Staff: Mentor

If the field is uniform (e.g. between two parallel oppositely-charged plates), the net force on the induced charge is zero. If the field is not uniform, then you do get a small force. Suppose the field comes from a small positively charged object held near the BB. The side of the BB nearest the object gains a - charge and the opposite side gains a + charge. The - charge is slightly closer to the other object so the attractive force is stronger than the repulsive force, and there is a net attractive force.

This also happens with non-conducting objects. You can demonstrate this on a cold day with a comb and a bowl of Rice Krispies. Comb your hair to put a charge on the comb and hold it near the Rice Krispies.

7. Nov 26, 2008

### m.e.t.a.

Sorry simo, my oversight. For some reason I was imagining that you were trying to roll the BB around by tugging at it with some charged object. The above two responses are correct -- you do need a charge on the BB in order for it to feel a force from a non-divergent electric field. Have you tried touching the BB to one of the plates while they are charged? If the voltage is high enough, and the friction low enough, the BB should begin to oscillate between the plates, ferrying charge as it does so.

8. Nov 26, 2008

### Redbelly98

Staff Emeritus
jtbell is right, a nonuniform E-field is needed. Polarizable objects are then attract to regions of stronger field.

This has been used to trap atoms with lasers, and also levitate ~micron-sized dielectric spheres. Googling "optical tweezers" would likely produce many hits. But to move something as large and heavy as a copper BB may not be practical.

One could probably calculate the E-field gradient required to produce a force equal to the weight, making some simple assumptions about the charge distribution on the BB.

EDIT:
The concentration of conduction electrons in copper is 8.5 × 10^28 per m^3, according to 2 sources I found in a google search.

Last edited: Nov 26, 2008
9. Nov 26, 2008

### simo

If the method of images is applied to a conducting sphere and a point charge, it can be shown that the two will attract. Therefore, I will try charging a single plate and getting rid of the other.

10. Nov 26, 2008

### Redbelly98

Staff Emeritus
I'm pretty sure that a 9V battery will not produce anywhere close to the field strength required to move a BB. I'm going to try the calculation I outlined in Post #8, and will post here with the result if I can get it to work.

11. Nov 26, 2008

### Redbelly98

Staff Emeritus
I'm getting an E-field of about 107V/m, with a gradient of 3x109V/m2, to produce a lifting force for a 3mm-radius copper sphere.

When I have time I'll try to post the details of the calculation, but I will be away from the computer most of tonight and then going out of town for a few days.

12. Nov 26, 2008

### simo

Cool, thanks.

13. Nov 27, 2008

### Naty1

Seems to me you either need a magnetic material or a charged one, to interact with a magnetic or an electric field, respectively.

Think it could work with a piece of wood? or plastic? Then the material matters!

Is the bb solid copper or steel coated copper? Is copper magnetic? How does a copper coating block electromagnetic fields from the interior?

Would you be willing to try a coil of wire..... maybe wrapped around a pencil?? Just be sure it's long enough to have sufficient resistance so it doesn't short your battery. Move it relative to the bb...see if that helps

14. Nov 30, 2008

### Monocles

Well if thats the case then you aren't going to be able to do this in normal air since electricity discharges at 3x10^6 V/m in normal air.

15. Nov 30, 2008

### Redbelly98

Staff Emeritus
Okay, here it is.

EDIT: this is not an exact calculation, but rather an estimate of the electric field strength needed to move a copper BB.

A copper sphere of radius r is placed in an external electric field. Let
Eo = the value of the external field at the sphere's center
and
dE/dx = the gradient of the E-field

First we use the average field Eo to estimate how much charge, +q and -q, is induced on the sphere.
After we estimate q, we'll use dE/dx to estimate the net force on the +q & -q charges.

Since this is just a rough estimate, imagine that the +q & -q induced charges:
• each make a disk of radius rq or area A
• are separated by a distance dq within the sphere
The field from the charges should be equal in magnitude to Eo, in order to produce 0 E-field within the conducting sphere. So

Eo = q / Aεo
or
q = Eo A εo

Since there is an E-field gradient, dE/dx, then let
Eoo = field at -q charge​
so that
Eoo +dq dE/dx = field at +q charge​
and the net force on the sphere is then

Fnet = -q Eoo + q [SIZE=+2]([/SIZE] Eoo + dq dE/dx [SIZE=+2])[/SIZE]
= q dq dE/dx ​
= Eo A εo dq dE/dx ​

Equating this force with the sphere's weight:

Eo A εo dq dE/dx = m g
= ρ (4 π / 3) r3 g
so

$$E_o \ \frac{dE}{dx} = \frac{ \rho \frac{4 \pi }{ 3} r^3 g }{ A \epsilon_o d_q }$$

Let the +q and -q charged disks have a radius r/√2 so that
A = π r^2 / 2
and a separation dq=r√2
so that

\begin{flalign*} E_o \ \frac{dE}{dx} & = & & \frac{ \rho \frac{4 \pi }{ 3} r^3 g } { (\pi r^2 / 2) \epsilon_o r \sqrt{2}} \\ & = & & \frac{8}{3 \sqrt{2}} \rho g / \epsilon_o \\ & \approx & & 2 \rho g / \epsilon_o \end{flalign*}

To get some numbers out of this calculation, imagine that the field (which is Eo at the sphere's center) has a gradient that produces a field of (1/2)Eo at one side of the sphere and (3/2)Eo on the other side. In that case,

dE/dx = (3/2 - 1/2)Eo / 2r
= Eo / 2r​

and so

Eo^2 / 2r = 2 ρ g / εo
or
Eo = sqrt(4 r ρ g / εo)

Time to plug in numbers!

We have

r = 2.2 mm = 0.0022 m for a standard BB
rho = 9000 kg/m^2 for copper
g = 10 m/s^2
εo = 9e-12 C^2 / Nm^2

and so we get
Eo = 9 x 106 V/m
and
dE/dx = Eo / 2r = 2 x 109 V/m2

Good point.

Last edited: Dec 1, 2008
16. Nov 30, 2008

### simo

Thanks again Redbelly98, I guess I won't be building this device.

17. Dec 1, 2008

### m.e.t.a.

Simo, perhaps you could explain the purpose of the device you had envisaged? That way, people might offer suggestions of how you could redesign it so that it will achieve the same goal.

18. Dec 1, 2008

### simo

Well I didn't want to say because I didn't want to be scrutinized, but this would be a good opportunity for you all to explain how this device fails, if it fails. And you're not allowed to say the first law of thermodynamics because that's too obvious.

What I'm trying to design is a perpetual motion device. See the attachment. Basically a neutral conducting sphere is drawn up a neutral insulated ramp by an electric field. At the top of the ramp, it falls into a conducting bucket which then allows the ball to returns to the bottom. The point of the buchet is to eliminate the e-field.

This would be in a vacuum

Note: Everything that is grey is a conductor. The charged plate goes no higher than the tip of the ramp.

File size:
31 KB
Views:
110