# I Does it matter where to connect a conductor to ground?

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1. Oct 10, 2016

### greypilgrim

Hi.

If an electrically neutral, conducting rod is brought close to a (say negatively) charged object with one end, charges will separate due to electrostatic induction roughly as follows:

Let's now connect the rod to the (far-away) ground with a long cable. Does it make a difference whether we connect the cable at points A, B, or C? How does the charge picture qualitatively look like afterwards?

2. Oct 10, 2016

### lychette

It does not matter where you make the connection to the rod. The negative charges will flow to ground (Earth) leaving the rod + charged

3. Oct 10, 2016

### greypilgrim

But why? Suppose we connect the cable at C. Why should the electrons move to this point, against the repelling Coulomb force from the sphere?

4. Oct 11, 2016

### Svein

As usual: It depends. Is the other object connected to ground? Is it close to ground?

Since you are conducting a thought experiment, I will give you the textbook answer: No, it does not matter (but in a thought experiment the ground is usually infinitely far away). Since there is no conducting path between the objects, there is no reason for electrons to move.

5. Oct 11, 2016

### greypilgrim

No and no. It might even be a charged insulator, I'm only interested in the conducting rod.

I'm talking about the excess electrons on the left side of the rod. If I connect the cable at A, it's clear that they will move further away from the sphere and into the ground. But if I connected the cable at C, they'd first have to move closer to the sphere. Why should they do this? What force drives them there?

6. Oct 11, 2016

### lychette

The key to this is to realise that the rod is a conductor and therefore will be at the same POTENTIAL. The -ve ball wants to make -ve charges on the rod flow to earth. Because the charges on the rod are at the same potential it does not matter where the connection to earth is made, the negative charges on the rod will flow to earth. If the connection with earth is removed and then the -ve ball is removed the rod will be +ve charged.
This is called charging by induction. Understanding what is meant by 'potential' is the key.

7. Oct 12, 2016

### davenn

huh ? same potential as what ?

I think you are missing the point that there will be charge separation along the length of the rod because of the influence of the charged sphere and that is what the OP is referring to

8. Oct 12, 2016

### CWatters

If we were considering -ve charges in free space (no conductor present) then yes, some force would be required to move them closer to the charged sphere. That's because the electric field created by the charged sphere extends out into space.

However introducing the conducting rod changes the field pattern created by the charged sphere. An electric field cannot penetrate a conductor so there is no field inside it. If there is no field inside the conductor then all the charges are the same "electrical distance" from the sphere. No force is required for them to move up and down the conductor in either direction.

9. Oct 12, 2016

### CWatters

I think this is roughly how the field ends up... The field cannot penetrate the conductor so it bends around it.

10. Oct 12, 2016

### greypilgrim

What kind of lines are those? Obviously not electric field lines, are they equipotential lines?

11. Oct 12, 2016

### CWatters

Yes sorry, equipotential lines.

One nearest to the sphere should really be more circular.

12. Oct 13, 2016

### lychette

I will reply to you again....you need to have an appreciation of POTENTIAL in the behaviour of electric charges. I will summarise for you:
1) The -ve ball is at a -ve potential (can be worked out if charge and diameter of ball are known)
2) ground/earth is at a potential of zero
3) The conducting rod is at a potential (-ve) somewhere below zero but above that of the ball
4) because it is conducting charges can flow over the rod, this ensures that all points on the rod are at the same potential....you have a simplified diagram showing how the charges might be displaced. (do you realise that the only charges that move are -ve electrons? the +ve charges are the nuclei of fixed atoms. ....not too important but worth remembering)
5) when the rod is connected to earth the whole rod is now at zero potential....this is the crucial point...the -ve charges (electrons) on the rod will be repelled by the +ve ball to ground....you could say that they are attracted to ground (which is zero potential!)
6) the negative charges are lost from the rod which is therefore + charged when the connection to ground is removed.
IT DOES NOT MATTER WHERE, ON THE ROD, THE CONNECTION TO GROUND IS MADE....ALL POINTS ON THE ROD ARE AT THE SAME POTENTIAL !!!

i WILL NOT MAKE ANY MORE CONTRIBUTIONS TO THIS POST....i HOPE SOMEONE GAINS SOMETHING POSITIVE FROM THIS.
Any criticisms of the physics in my post is most welcome

13. Oct 14, 2016

### Staff: Mentor

A rather heated exchange has been deleted, and the thread is reopened.

14. Oct 14, 2016

### Staff: Mentor

The repelling force from the sphere is exactly counteracted by the attracting force from the positive charges on the conductor surface (both the rod and the ground). There is nothing opposing the motion. That is exactly why the conductor is considered equipotential.

Last edited: Oct 14, 2016
15. Oct 14, 2016

### greypilgrim

Does this mean my picture was already wrong in the first place (without connection to ground)? Why should the negative charges accumulate on the left side of the rod if they're free to move?

And why do those two forces exactly cancel? Charges normally can't leave the surface of a charged conductor because "normal forces" hold them back. Why can't the repelling force from the sphere be stronger than the attractive one from the positive charges on the rod, and the difference be compensated by those "normal forces"?

16. Oct 14, 2016

### Staff: Mentor

The picture is fine as long as you recognize that the charges indicated are surface charges.

Because if they didn't then there would be currents and it would not be an electrostatic situation.

In the electrostatic scenario the surface charges adjust so that the E field outside the conductor is normal to the surface and the E field inside the conductor is 0.

17. Oct 14, 2016

### greypilgrim

If the repelling force was stronger, there would be no currents because the electrons on the left can't leave the surface.
Yes, but in this case the E-field inside the conductor has three contributions: From the negative charge on the sphere, the positive charge on the right side of the rod and the negative charge on the left side of the rod. In #14 you say that already the forces from the sphere and the right side of the rod cancel.

18. Oct 14, 2016

### Staff: Mentor

I am sorry for any misunderstanding. In #14 I should have referred to the total contribution from all charges. I wouldn't separate this problem into 3 contributions.

Last edited: Oct 14, 2016
19. Oct 14, 2016

### CWatters

Perhaps worth repeating...

When the electrons flow away from the sphere they create a field in the opposite direction to that created by the sphere. So the net field is reduced. They keep flowing until the net field is zero and you have the electrostatic situation.

20. Oct 14, 2016

### greypilgrim

I think I gained some insight that allows me to formulate my question (hopefully) more clearly.

If the rod is grounded, it will become positively charged. From my picture I conclude that this means the negative excess charges on the left move into the ground. I agree that the field inside the rod is zero and charges inside the rod are therefore free to move wherever they want. But the excess charges I'm talking about are surface charges, and the electric field is in general not zero at the surface of a conductor. In this particular case, on the left end of the rod there are field vectors normal to the rod surface and pointing inside. This means the negative surface charges on the left feel an electric force directed outside, normal to the surface. This force is counteracted by some "surface force" that prevents charges from leaving the conductor under normal circumstances.

As far as I can see it's simply not true that those excess charges on the surface feel no electric field and hence can move into the ground when a grounded cable is connected at C.