# A few questions regarding Neutral Objects

1. Homework Statement

Hello,
I'm learning electricity and I'm having a few conceptual questions regarding the subject ( especially about neutral objects ) which I'm unsure of the answers and I'd be happy if someone could help me:

1. Is the charge density of a neutral object ( doesn't matter if it's a conductor or not ) always zero? If so, then if for example I have a conducting neutral wire ( not charged ) and I connect it to a dc battery , then does the wire get charged? does its charge density change?

2. If an object is neutral , then does that mean it doesn't exert a potential at a point? (i.e: it exerts only a zero potential at distance 'r' : $V(r) = 0$ ) , if so, then does this mean that only a charged object can exert a potential at a point? (i.e: it exerts only a potential that is different from zero at distance 'r' : $V(r) = k*\frac{Q}{r}$).

3. If I have a neutral object which is a conductor , then If I put that object in an electric field , the object un-neutralizes and gets charged ( and therefore charge density changes ) - why does that happen?

## Homework Equations

- $Q = \sigma * A$
- Gauss's Law
- $V(r) = k*\frac{Q}{r}$
- $E(r) = k*\frac{Q}{r^2}$

## The Attempt at a Solution

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BvU
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Hi CG&C,

1. Averaged over the entire object, yes.
A battery does send a very small amount of charge into a connected wire. Very, very small (remember this; once you can do some quantitative calculations you should try to make an estimate).
2. No, it does not mean that. Simple example: two charges a distance apart constitute a neutral object, but there is a non-zero potential field (a so-called dipole field).
3.
If I have a neutral object which is a conductor , then If I put that object in an electric field , the object un-neutralizes
It does not. Where did you get that (exact formulation, please) ?
What can be said is that the charges in the object re-distribute themselves in a very specific manner, namely in such a way that the entire surface of the conductor is at the same potential.

3. Here your post is severely lacking (and in fact I'm not even allowed to assist, according to the PF rules). Do try to write down an attempt at solution -- it really helps you help yourself (and us to help you better, instead of just chewing out the 'truth').

Hi CG&C,

1. Averaged over the entire object, yes.
A battery does send a very small amount of charge into a connected wire. Very, very small (remember this; once you can do some quantitative calculations you should try to make an estimate).
2. No, it does not mean that. Simple example: two charges a distance apart constitute a neutral object, but there is a non-zero potential field (a so-called dipole field).
3.
It does not. Where did you get that (exact formulation, please) ?
What can be said is that the charges in the object re-distribute themselves in a very specific manner, namely in such a way that the entire surface of the conductor is at the same potential.

3. Here your post is severely lacking (and in fact I'm not even allowed to assist, according to the PF rules). Do try to write down an attempt at solution -- it really helps you help yourself (and us to help you better, instead of just chewing out the 'truth').
What formulas refer to this calculation? Ohms law?

Since you said that neutral charge can actually exert a potential field , then does this mean a simple conducting neutral wire always exerts a potential field? , if so , then why do I need a battery to create an external potential field in order to move charges in an electric circuit? I have the potential field from the wires themselves ( without battery ) and that's enough.

Well, my idea was that since the charges in the object re-distribute themselves when the conducting neutral object is put into an electric field, then , it's charge density changes, therefore the average charge density is no longer zero and therefore , the object now has charge ( since it has non zero charge density ) by looking at the equation: $Q = \sigma*A$

haruspex
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Since you said that neutral charge can actually exert a potential field , then does this mean a simple conducting neutral wire always exerts a potential field?
A body with uniformly neutral charge will not give rise to a field.
Any non-uniform charge distribution will give rise to a field. If that body has net neutral charge it could be an insulator or a conductor subjected to an external field. In the latter case, the induced charge distribution will give rise to a field which exactly cancels the external field within the conducting body.

A body with uniformly neutral charge will not give rise to a field.
Any non-uniform charge distribution will give rise to a field. If that body has net neutral charge it could be an insulator or a conductor subjected to an external field. In the latter case, the induced charge distribution will give rise to a field which exactly cancels the external field within the conducting body.
But what about a dipole? It can be seen as a body with uniformly neutral charge that does rise a potential field.

... two charges a distance apart constitute a neutral object, but there is a non-zero potential field (a so-called dipole field).

haruspex
Homework Helper
Gold Member
But what about a dipole? It can be seen as a body with uniformly neutral charge that does rise a potential field.
Not quite uniform.

Not quite uniform.
Why isn't the charge distribution of a dipole uniform? how do you measure the uniformity in this case?

And when you said:
... In the latter case, the induced charge distribution will give rise to a field which exactly cancels the external field within the conducting body.
Did you mean the electric field inside a conducting body or potential field?

haruspex
Homework Helper
Gold Member
Why isn't the charge distribution of a dipole uniform? how do you measure the uniformity in this case?
There is a small separation between the opposite charges, so not uniform.
Did you mean the electric field
Yes.

BvU
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2019 Award
Since you said that neutral charge can actually exert a potential field
We are in a physics forum. You should read more carefully and quote a lot more carefully. A neutral object is not a neutral charge.

BvU
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2019 Award
What formulas refer to this calculation? Ohms law?
No. To determine if an object is neutral, all you have to do is add up all the charge that is present. If that comes out at exactly zero, the object is neutral.
the charges in the object re-distribute themselves
They do, but they do not leave the object, so the total charge does not change.
the object now has charge ( since it has non zero charge density )
Your logic is incorrect, as was already explained to you: non-zero charge density does not mean non-zero charge.

Your logic is incorrect, as was already explained to you: non-zero charge density does not mean non-zero charge.
So let me get this straight by a formula:
Since the charge $Q$ of an object with charge distribution $\sigma(x,y)$ ( not neccesarily uniform charge density ) can be found with the following formula: $Q = \iiint \sigma(x,y) dxdy$

Then, for a neutral object: $Q = 0$ and then $0 = \iiint \sigma(x,y) dxdy$ , so from this I deduce that:
the charge density $\sigma(x,y)$ for a neutral object is not neccesarily zero , it's just that the sum of contributions of each infinitesimal area $dxdy$ multiplied by charge density $\sigma(x,y)$ at point (x,y) , is zero.

Would you say this deduction is correct?

In addition , the formula I was talking about in earlier post ( $Q = \sigma*A$ ) refered to average charge density, since $Q= 0$ then $0 = \sigma*A$
and therefore $\sigma = 0$ . but this $\sigma$ refers to the average charge density and not the specific charge density $\sigma(x,y)$ at point (x,y)

Would you say this deduction is also correct?

BvU
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An object with surface charge density $\sigma$ has surface charge $Q = \iint \sigma \, dA$ where the integration is over the entire surface.

An object with charge density $\rho$ has charge $Q = \iiint \rho \,dV$ where the integration is over the entire volume.

Would you say this deduction is correct?
In two dimensions, yes.

$0 = \sigma*A$
better use different symbols for $\sigma$ and $\sigma_{\rm avg}$

Would you say this deduction is also correct?
Yes

A body with uniformly neutral charge will not give rise to a field.
Any non-uniform charge distribution will give rise to a field. If that body has net neutral charge it could be an insulator or a conductor subjected to an external field. In the latter case, the induced charge distribution will give rise to a field which exactly cancels the external field within the conducting body.
I know that in nature the combination of opposite charges to create electrically neutral objects ( objects with total charge equal to zero) happens naturally without any outside intervention. So another question arises:

Do charges on neutral objects in nature tend to arrange uniformly ( without outside intervention ) so that the object does not give rise to any electric field nor potential field? ( since the charge arranges uniformly and since the object is neutral then the charge density of the object is zero and that implies that the object does not give rise to any field )

BvU
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I know that in nature the combination of opposite charges to create electrically neutral objects ( objects with total charge equal to zero) happens naturally
Where do you get such all-knowing knowledge ? And what does it apply to ? ('electrically neutral objects' can range from elementary particles to mountain ranges, or galaxies or bigger). 'naturally'?

Where do you get such all-knowing knowledge ? And what does it apply to ? ('electrically neutral objects' can range from elementary particles to mountain ranges, or galaxies or bigger). 'naturally'?
I meant macroscopic objects ( obviously ).
I'll show you an example to clarify:

If I'd put a neutral conducting wire near a closed circuit connected to a lamp as such:

Then, the potential at distance $r$ from the wire ( painted in black in the schematic ) with charge density $\lambda$ is given by the following equation: $V(r) \propto \lambda*log(r)$
Since there exists a potential at distance $r$ from the wire , it will allow charges to move in the circuit and therefore the lamp will turn on and shine ( resistance of lamp is approximately zero )
But obviously this does not happen in real life so my question is , why?

( so I brought up an 'all-knowing knowledge' from another source but this question was not addressed so I proposed the hypothesis that the charge density of the wire must be zero because only that way the wire does not give rise to any fields and therefore the lamp wouldn't turn on and this proposal seems to fit with reality ( macroscopically at least) because otherwise , lots of things would turn on without them being connected physically to a voltage source, but again , this hypothesis might be false , hence the question )

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BvU
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it will allow charges to move in the circuit
And in which direction would that be ?

How far along are you in your study of electricity and magnetism ?

And in which direction would that be ?

How far along are you in your study of electricity and magnetism ?
I assume the movement of charge would be in such direction ( red arrows show the direction ):

My reasoning is:
since positive charges move from higher potential to lower potential , then the charges in the circuit will flow as shown. This is because as we move further away from the wire the potential increases but as we move closer to the wire the potential decreases ( I think the formula for the potential is wrong because the potential should decrease as we move further away from the wire ,but it doesn't really matter for the current answer ) .

So there exists a charge flow in the circuit, but such charge flow won't turn the lamp on because the total current in the circuit will be zero ( since the superposition of two currents with equal magnitude but different directions will give zero ) ...

Ok then, in this specific example the total current in the circuit is zero..
but if there was a different neutral object ( not neccesarily a wire ) creating a potential field such that the total current in the circuit is not zero? could such situation even exist? if it does, then how come electric stuff dont just get turned on randomly at real life? is it perhaps because the current\voltage is too low?

And i'm just learning about potentials and simple DC circuit analysis.

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BvU
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higher potential to lower potential
What potential ? I must assume you mean the potential from the
charged ($\lambda$) , not neutral
I'd put a neutral conducting wire
conducting wire ? yes that makes the charge carriers in the loop rearrange themselves, in such a way that the field inside the conducting loop is zero. That means: only a very little bit, and only during an extremely short time after the $\lambda$ is turned on.

( I think the formula for the potential is wrong because the potential should decrease as we move further away from the wire ,but it doesn't really matter for the current answer )
Kudos for your intuition. Where did you get the expression and what assumptions accompanied it ?

but if there was a different neutral object ( not neccesarily a wire ) creating a potential field such that the total current in the circuit is not zero? could such situation even exist?
Again, kudos for imaginative thinking. Note that it is not potential that makes charge move, but electric field (the change in potential) (*). It should lead to a conclusion that in electrostatics you only get a steady current in a circuit if you can go around and manage to end up at a different potential than what you started with. Which is 'difficult' since there is only one single potential at a given point. So electric machinery doesn't start moving spontaneously .... (fortunately).

On the other hand, electric machinery is known to function in a reasonably reliable manner. How come ? Well, it turns out that (*) magnetic fields can exert forces on moving charge carriers and that helps a lot. But now we dwelll outside the scope of electrostatics and of this thread...

How far along are you in your study of electricity and magnetism ?