Ways of transferring charge between objects (conduction and induction)

[MatinSAR]

Homework Statement

We have two similar conducting spheres. The charge of sphere b is equal to 30 and the charge of a is equal to -10 (In SI unit). We bring the plastic rod clos to sphere a. We close the k1 key and open it again. Then we close and open key 2. The final charge of two spheres?

Relevant Equations

-

Two ideas:
1) Negative charged object get close to A then we close K1 key. So -10SI charge from sphere A moves to sphere B and 30SI charge moves from sphere B to sphere A. At this moment sphere A has 30SI charge and sphere B has -10SI charge.
2) We close K1 key so system of 2 spheres should have $30-10=20$SI charge. But because of plastic rod that is close to sphere A all this 20SI charge moves to A and B doesn't have any charge.

I cannot find out which one is correct to continue ...

 

[PhDeezNutz]

Does the problem statement specify if the plastic rod is positive or negative (I couldn't find it in your original post) or maybe even perhaps neutral so there is a polarizing effect?

 

[MatinSAR]

Does the problem statement specify if the plastic rod is positive or negative (I couldn't find it in your original post) or maybe even perhaps neutral so there is a polarizing effect?

Is it possible for plastic to have positive charge?
Anyway suppose the plastic rod has negative charge.

 

[Steve4Physics]

Homework Statement: We have two similar conducting spheres. The charge of sphere b is equal to 30 and the charge of a is equal to -10 (In SI unit). We bring the plastic rod clos to sphere a. We close the k1 key and open it again. Then we close and open key 2. The final charge of two spheres?
Relevant Equations: -

View attachment 337778

The problem refers to 'the plastic rod' but this has not been mentioned before and is not shown in the diagram. Is the question/diagram complete?

By the way, the SI unit of charge is the coulomb. If you have a charges of -10C and +30C on the spheres these are somewhat large charges. Expect something like this and complaints from your neighbours! Maybe you mean units of microcoulombs, $\mu C$.

 

[MatinSAR]

The problem refers to 'the plastic rod' but this has not been mentioned before and is not shown in the diagram. Is the question/diagram complete?

It only mentioned that we bring that negative charged plastic rod close to sphere A. So we can add it to the diagram if we want to ...

By the way, the SI unit of charge is the coulomb. If you have a charges of -10C and +30C on the spheres these are somewhat large charges. Expect something like this and complaints from your neighbours! Maybe you mean units of microcoulombs, $\mu C$.

lol ... I will try to do the experiment outside of the town ...
We can forget about the units. The charges can be -10e or 30e ... But that's not important I should find 2 numbers as answer ...

Edit:
Yes the charges are -10$\mu C$ and 30$\mu C$.

 

[kuruman]

Is this a question that you found in a book or did you put it together yourself? I am asking because a numerical answer cannot be found without the amount of charge and its polarity on the plastic rod. Even then, it would be difficult to get numbers out without doing some kind of numerical computation.

That said, please explain why should charge move to and from conductors and the Earth? When such flow is occurring, what must be true for it to stop?

 

[MatinSAR]

Is this a question that you found in a book or did you put it together yourself? I am asking because a numerical answer cannot be found without the amount of charge and aaaaa on the plastic rod. Even then, it would be difficult to get numbers out without doing some kind of numerical computation.

It is from a book which is not in English. I've thought that we need to know about the amount of charge on the plastic rod. Then the idea came to my mind that maybe we can consider the negative charge of the rod big enough so that it can pull all the positive charges towards itself.

aaaaa

I did not understand this part.

That said, please explain why should charge move to and from conductors and the Earth? When such flow is occurring, what must be true for it to stop?

When we close K1 key because of that negative charged rod the negative charges on sphere A want to get away from the rod so they move to the furthest point sphere B.

 

[MatinSAR]

If we assume only negative charges move:
In this case, only 10 microcoulombs of negative charge will go from A to B, so A will be neutral and B will have a charge of 10-30=20 microcoulombs.

I think this is better that my first two ideas ...

Edit:
I've always hate this part of E&M because I always find more than 1 answer for any question like this.
Anyway if this questions seem untrue and worthless please let me know I will skip it.

 

[Steve4Physics]

If we assume only negative charges move:
In this case, only 10 microcoulombs of negative charge will go from A to B, so A will be neutral and B will have a charge of 10-30=20 microcoulombs.

It doesn't work like that. Only some of A's negative charge would move to B. Exactly how much would depend on the charge (amount and distribution) on the rod. Even if you knew this, it would be a complex calculation (as already noted by @kuruman).

Anyway if this questions seem untrue and worthless please let me know I will skip it.

In my opinion you should skip it!

 

[MatinSAR]

In my opinion you should skip it!

I will. Thanks for your help.

 

[kuruman]

You can answer this question with numbers if you get rid of the charged rod. Just close and open K1 and then close and open K2. How much charge would be on each conductor then?

 

[kuruman]

I did not understand this part.

It was a typo that I saw and fixed but not fast enough for you not to notice.

 

[MatinSAR]

You can answer this question with numbers if you get rid of the charged rod. Just close and open K1 and then close and open K2. How much charge would be on each conductor then?

But this will change the question completely. Without that rod it's obvious that after doing the tasks you've mentioned we have no charge on sphere B and we have $10 \mu C$ charge on sphere A.

 

[Steve4Physics]

You can answer this question with numbers if you get rid of the charged rod. Just close and open K1 and then close and open K2. How much charge would be on each conductor then?

I'm not sure it's that simple.

Closing and opening K1 will leave 10$\mu C$ on each sphere.

Then closing K2 then grounds sphere B, making its potential zero. But I believe that does not leave a zero charge on B. There will be an induced 'mirror' charge on B so that B is at ground potential in the presence of A's field. Finding the size/distribution of the mirror charge on B is not trivial.

Finally, opening K2 will have no further effect.

 

[MatinSAR]

Then closing K2 then grounds sphere B, making its potential zero. But I believe that does not leave a zero charge on B. There will be an induced 'mirror' charge on B so that B is at ground potential in the presence of A's field. Finding the size/distribution of the mirror charge on B is not trivial.

I haven't read about mirror charges anywhere ...
The closest thing I can think of "a mirror charge" is the method of image charges which helps us find potential without solving $\nabla ^2 \phi = 0$ in electrostatics. And I cannot see how it relates to this question ...

Edit :

I really dont want to repeat my mistake and I do not want to waste anyone's time with a bad question. As you said it is better to skip it ...

@PhDeezNutz @Steve4Physics @kuruman Thanks for your time.

 

[Steve4Physics]

I haven't read about mirror charges anywhere ...
The closest thing I can think of "a mirror charge" is the method of image charges which helps us find potential without solving $\nabla ^2 \phi = 0$ in electrostatics. And I cannot see how it relates to this question ...

For information...

An example: you put a point charge, +q, a distance d in front of an infinite conducting plane. The resulting field behaves as if there were no plane, but just two charges, +q and its reflection (-q) separated by 2d.

Of course, the -q charge doesn't exist. The resulting field is actually the sum of the fields from the +q charge and from induced negative surface charges on the conducting plane.

 

[kuruman]

I haven't read about mirror charges anywhere ...

You don't need to know about mirror charges. At the end of the procedure that I described in post #11 the charge on conductor B will be the same as in the following procedure:
1. You open both switches.
2. You put $10~\mu\text{C}$ on conductor A
3. You close K2.

Do you see now why there will be charge on B?

 

[MatinSAR]

For information...

An example: you put a point charge, +q, a distance d in front of an infinite conducting plane. The resulting field behaves as if there were no plane, but just two charges, +q and its reflection (-q) separated by 2d.

Of course, the -q charge doesn't exist. The resulting field is actually the sum of the fields from the +q charge and from induced negative surface charges on the conducting plane.

I call this the method of image charges ... intersting ! I didn't notice the relation. Thanks @Steve4Physics ...

 

[MatinSAR]

You don't need to know about mirror charges. At the end of the procedure that I described in post #11 the charge on conductor B will be the same as in the following procedure:
1. You open both switches.
2. You put $10~\mu\text{C}$ on conductor A
3. You close K2.

Do you see now why there will be charge on B?

Not really.
In post #11 you've stated that we close and open K1 so there is equal charge on each sphere. Both spheres should have $\dfrac {30-10}{2} = 10 \mu C$ charge. Then we close and open K2. It doesn't change charge of sphere A. But this action grounds sphere B so it should have 0 charge if I ignore the point about image charges ...

 

[kuruman]

if I ignore the point about image charges ...

You can't because of the proximity of charged sphere A. Electrons from the Earth will be attracted and flow into sphere B. That's the first step in charging by induction.

 

[MatinSAR]

You can't because of the proximity of charged sphere A. Electrons from the Earth will be attracted and flow into sphere B. That's the first step in charging by induction.

But that 2 spheres are not connected and they are not too close to each other. Does sphere A have effect on sphere B even if I suppose they are far from each other?

To answer your question I would say that if sphere A has charge of Q then sphere B should have charge of -Q.

 

[Steve4Physics]

I was wrong (or at least misleading) when in Post #14 I said “Finding the size … of the mirror charge on B is not trivial.”

It is trivial in ideal circumstances.

In the ideal case, the only charge-carrying objects in the neighbourhood are A and B. We can neglect the charge-carrying ability of the ground itself by assuming that the dimensions of A and B, and their separation, are very small compared to the distance to the ground.

Every field-line must start and end on a charge.

In this situation, every field line leaving A must end on B - there's nowhere else for the field line to go. It follows that the induced ('mirror') charge on B is equal and opposite to A's charge.

So under ideal conditions, B’s final charge is -10$\mu C$.

I presume this is what @kuruman has in mind (?).

 

[kuruman]

You don't need to connect the spheres. In the simplest model for a conductor electrons are free to move inside it in response to external electric fields. A has lost electrons and has therefore net positive charge while B has zero net charge. Charged conductor A creates an electric field through B that causes some of the free electrons in B to redistribute themselves so that the electric field inside B is zero. A must be infinitely far away from B in order not to have an effect on it.

The presence of positively-charged A raises the electrostatic potential in its vicinity. The Earth is at zero potential by definition. This means that there is a potential difference between B and the Earth. When B is connected electrically to the Earth by closing K2, B becomes an extension of the Earth and must also be at zero potential. Negatively-charged electrons will flow from the Earth to B until its potential is lowered to zero. This will change the net charge on B from zero to negative. When K2 is reopened, the negative charges are trapped on B. We say that the originally neutral B has be charged by induction to carry net negative charge.

 

[kuruman]

Every field-line must start and end on a charge.

True, but this doesn't imply that all the lines leaving A will end at B. Lines leaving A can also end at charges at infinity or charges induced on the surface of the Earth which is part of the picture.

 

[MatinSAR]

So under ideal conditions, B’s final charge is -10$\mu C$.

That's what I'm thinking ...

A must be infinitely far away from B in order not to have an effect on it.

Agree. I was wrong.

Negatively-charged electrons will flow from the Earth to B until its potential is lowered to zero. This will change the net charge on B from zero to negative.

Can't we find the amount of charge as I said in post #21?

 

[Steve4Physics]

True, but this doesn't imply that all the lines leaving A will end at B. Lines leaving A can also end at charges at infinity or charges induced on the surface of the Earth which is part of the picture.

For the ideal case (Post #22) all field lines leaving A will necessarily end on B.

But I agree with you that in a real-world situation, this will not be the case due to induced charges on remote objects and on the earth itself.