Redbelly98 said:
When I have time I'll try to post the details of the calculation ...
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
E
o = 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 E
o 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 E
o, in order to produce 0 E-field within the conducting sphere. So
E
o = q / Aε
o
or
q = E
o A ε
o
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
F
net = -q E
oo + q [SIZE=+2]( E
oo + d
q dE/dx [SIZE=+2])
= q dq dE/dx
= Eo A εo dq dE/dx
Equating this force with the sphere's weight:
E
o A ε
o d
q dE/dx = m g
= ρ (4 π / 3) r3 g
so
<br />
E_o \ \frac{dE}{dx} = \frac{ \rho \frac{4 \pi }{ 3} r^3 g }{ A \epsilon_o d_q }<br />
Let the +q and -q charged disks have a radius r/√2 so that
A = π r^2 / 2
and a separation d
q=r√2
so that
<br />
\begin{flalign*}<br />
E_o \ \frac{dE}{dx} & = & & \frac{ \rho \frac{4 \pi }{ 3} r^3 g } { (\pi r^2 / 2) \epsilon_o r \sqrt{2}} \\<br />
& = & & \frac{8}{3 \sqrt{2}} \rho g / \epsilon_o \\<br />
& \approx & & 2 \rho g / \epsilon_o<br />
\end{flalign*}<br />
To get some numbers out of this calculation, imagine that the field (which is E
o at the sphere's center) has a gradient that produces a field of (1/2)E
o at one side of the sphere and (3/2)E
o on the other side. In that case,
dE/dx = (3/2 - 1/2)E
o / 2r
= Eo / 2r
and so
E
o^2 / 2r = 2 ρ g / ε
o
or
E
o = sqrt(4 r ρ g / ε
o)
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
ε
o = 9e-12 C^2 / Nm^2
and so we get
E
o = 9 x 10
6 V/m
and
dE/dx = E
o / 2r = 2 x 10
9 V/m
2
Monocles said:
Well if that's 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.
Good point.
