Electrical potential of a thin wire in an E field

[coquelicot]

Assume that an infinite metallic plate A lies in the xy-plane, and another infinite metallic plate B is parallel to A and at height z = h.
The potential of plate A is 0, and the potential of plate B is constant and equal to V.
So, there is a uniform electrostatic field E between plates A and B.
Between the plates, but without touching them, there is a very thin copper wire of length L (L < h).
It can be supposed that the wire is located on the z axis, between z = a and z = b (so b-a = L).

My question: what is the potential of the copper wire (it is known that it is constant along the wire, since the wire can be seen as an ideal conductor).

Edit: it is not true that the electric field is uniform since the copper wire modify it. I meant "the electric field is constant before the copper wire is introduced".

 

[jartsa]

By symmetry we know that a charge that leaves the wire at the middle point of the wire and then moves straight away from wire does not lose or gain potential energy.

So, if we put positive test charges near the ends of the wire and at the middle of the wire, and then remove the wire, one test charge gains energy, one test charge loses energy, and the energy of the test charge at the middle stays the same.

(Any conducting material nearby would destroy the symmetry, so I hope the E-field was created by charged non-conducting plates)

 

[coquelicot]

In what way does that answer the question ?

N.B: what symmetry? (I've not said the wire is in the middle of the space between the plates).

 

[jartsa]

In what way does that answer the question ?

N.B: what symmetry? (I've not said the wire is in the middle of the space between the plates).

The test charge moving away from the middle of the wire moves so that its distance to the charged non-conducting plates does not change. The motion is symmetric in that way.

And then the test charge is neither attracted or repelled by the wire, because of symmetry. I mean, moving the charge away from the wire does not require any work, so the potential of the charge does not change.

 

[coquelicot]

Sorry, I see no answer here, and even not the smallest hint. In fact, I see no relation of your answer with my question.

 

[Nugatory]

And then the test charge is neither attracted or repelled by the wire, because of symmetry.

I may be misunderstanding what you're saying here...
The symmetry doesn't mean that there will be no force on a test particle moving (in the plane parallel to the plates and midway between them) away from the wire, but only that if there is a force it will be same in all directions.

 

[PeroK]

Summary:: potential of a thin wire in an E-field between two plates, the wire being parallel to E.

Assume that an infinite metallic plate A lies in the xy-plane, and another infinite metallic plate B is parallel to A and at height z = h.
The potential of plate A is 0, and the potential of plate B is constant and equal to V.
So, there is a uniform electrostatic field E between plates A and B.
Between the plates, but without touching them, there is a very thin copper wire of length L (L < h).
It can be supposed that the wire is located on the z axis, between z = a and z = b (so b-a = L).

My question: what is the potential of the copper wire (it is known that it is constant along the wire, since the wire can be seen as an ideal conductor).

Edit: it is not true that the electric field is uniform since the copper wire modify it. I meant "the electric field is constant before the copper wire is introduced".

You could model the wire as a thin cylinder and try to solve Laplace's equation.

Or, maybe you could use the method of images?

 

[coquelicot]

You could model the wire as a thin cylinder and try to solve Laplace's equation.

Or, maybe you could use the method of images?

Yes, that is why I tagged my question with "Laplace equations". But it would be nice if someone has already solved them in this particular case (I'm not sure I have sufficient background in DPE to solve them).

 

[jartsa]

Sorry, I see no answer here, and even not the smallest hint. In fact, I see no relation of your answer with my question.

Sorry, I was solving a problem with non-conducting charged plates, because I didn't read the question which clearly says the plates are mede of metal.

 

[rude man]

I think the exact solution is beyond any elementary math.

We want to find the two E fields right under & above the wire ends (lines perpendicular to the plates A and B from a and b).

Say we start with the needle exactly halfway between the plates. Then the E fields will be equal and easily computed.
Now we move the needle a distance $\delta a$ away from A and toward B. We now assume unequal E fields $E_a$ and $E_b$:

$E_a = E - \delta E_a$ and $E_b = E + \delta E_b$. Note that $\delta E_a$ and $\delta E_b$ are assumed positive here. $\delta a$ can be any finite number within physical bounds.

With some algebra it can be shown that the condition

$$ \frac {\delta E_b - \delta E_a} {\delta E_b + \delta E_a} =\frac { \delta a} {a} $$

will be met. So for $\delta a << a, \delta E_a = \delta E_b = 0$ and the potential is easily calculated.
Unfortunately, as I said, separating $E_a$ and $E_b$ for all positions of the wire seems out of scope.

 

[coquelicot]

Say we start with the needle exactly halfway between the plates. Then the E fields will be equal and easily computed.

They will be equal by symmetry (and the potential will be equal to V/2 by symmetry), but I don't see how they can be easily computed.

Now we move the needle a distance $\delta a$ away from A and toward B. We now assume unequal E fields $E_a$ and $E_b$:

$E_a = E - \delta E_a$ and $E_b = E + \delta E_b$. Note that $\delta E_a$ and $\delta E_b$ are assumed positive here. $\delta a$ can be any finite number within physical bounds.

With some algebra it can be shown that the condition

$$ \frac {\delta E_b - \delta E_a} {\delta E_b + \delta E_a} =\frac { \delta a} {a} $$

will be met.

?

So for $\delta a << a, \delta E_a = \delta E_b = 0$ and the potential is easily calculated.

? ?

 

[rude man]

They will be equal by symmetry (and the potential will be equal to V/2 by symmetry), but I don't see how they can be easily computed.

Let the E field between the wire ends and the respective plates when the wire is centered = $E$ and let the field far from the wire = $E_0$.

Then $\int_0^a E dl$ + $\int_b^h E dl$ = $\int_0^h E_0 dl.$
Do you see why?
Then evaluate the integrals & solve for E.
************************************************************************************

You could approximate by assuming that E goes down on one end as it goes up on the other by $\frac {\Delta E} {E} = - \frac {\Delta a} {a}.$

Then let
$V_{x1} = (E - ΔE)(a + Δa)$
$V_{x2} = (E + ΔE)(a - Δa)$
Then the error $\Delta V_x$ in computing the wire potential $$ V_x = \frac {V_{x1} + V_{x2}} {2} $$
is $\Delta V_x = - \Delta a \Delta E$ which is of 2nd order.