How Do Charges Separate on the Second Ball in Electrostatic Induction?

[vcsharp2003]

Homework Statement

Two metal coated pith balls are suspended by long insulating threads and touch each other as shown in diagram below. When a negatively charged ebonite rod is brought near one of the pith balls, they separate. Why does this separation occur?

Relevant Equations

$F= \dfrac {k q_1 q_2} {r^2}$

Electrostatic induction where charges on a conductor separate so that one end has +ve charges and the other end has -ve charges when a charged object is brought near the conductor.

I used the concept of electrostatic induction, which would cause the charges in metal ball near the ebonite rod to have +ve charges on end next to rod and a -ve charge on the end touching the other ball.

What confuses me is how charges separate on the second ball. The only way these balls can separate is if ball 2 has -ve charges on end touching the ball 1. ( ball 1 is next to ebonite rod)

What I get after applying electrostatic induction.

 

[Orodruin]

Your image only explains why the closer ball is attracted to the rod. Not why the far ball is repelled.

 

[vcsharp2003]

Your image only explains why the closer ball is attracted to the rod. Not why the far ball is repelled.

That's what I am unable to understand. Why would ball 2 be repelled towards the left?
I think ball 2 would have +ve charges move to end touching ball 1 and -ve charges move to the other end, and as a result ball 1 and ball 2 would now attract rather than repel.

 

[vcsharp2003]

Your image only explains why the closer ball is attracted to the rod. Not why the far ball is repelled.

I have updated the image of my attempt a moment ago, so you can see charges on both balls according to me. If I am correct then the balls would attract and not repel according to the charge distribution on these balls shown in the image of my attempt.

 

[haruspex]

That's what I am unable to understand. Why would ball 2 be repelled towards the left?
I think ball 2 would have +ve charges move to end touching ball 1 and -ve charges move to the other end, and as a result ball 1 and ball 2 would now attract rather than repel.

Think about all that happens before they separate. They are touching, remember.

 

[vcsharp2003]

Think about all that happens before they separate. They are touching, remember.

Would charges flow from one metal ball to the other metal ball and not simply separate on one metal ball?

 

[Orodruin]

Yes, if the balls touch charge will flow from one to the other since the balls are conductors.

 

[vcsharp2003]

Yes, if the balls touch charge will flow from one to the other since the balls are conductors.

Then I would get the situation as in image below.

Ball 2 could be either attracted or repelled from ball 1 based on vector sum of forces on ball 2.

 

[vcsharp2003]

If horizontal component of $F_2$ is large enough so it exceeds $F_1$ then only ball 2 would get repelled.

I guess we can assume so since question says ball 2 is repelled towards the left.

 

[haruspex]

If horizontal component of $F_2$ is large enough so it exceeds $F_1$ then only ball 2 would get repelled.

I guess we can assume so since question says ball 2 is repelled towards the left.

Maybe we can do better than assuming that.
If the charge on the rod is -Q and the induced charges are q and -q, is there a reason to suppose q <Q?

 

[vcsharp2003]

Maybe we can do better than assuming that.
If the charge on the rod is -Q and the induced charges are q and -q, is there a reason to suppose q <Q?

From symmetry of charges in electrostatics, that sounds true i.e charges on each metal ball should equal in magnitude the charge on ebony rod.

Then, the metal balls will attract each other with a very large force compared to force of repulsion between ebony rod and ball 2 due to very small distance between charges on metal balls.

This would mean that the metal balls don't separate.

 

[hmmm27]

What is the significance of "insulated threads" in the OP ?

 

[vcsharp2003]

What is the significance of "insulated threads" in the OP ?

So that charges do get conducted away and stay on the metal balls. Insulators do not allow charges to flow through it.

 

[Delta2]

Something tells me that the left ball ends up with charge -q and the right ball with charge +q, where q<<|Q|, Q the charge of the rod. That would explain it since their attractive force would be much smaller than the force from the rod on them. But I am not sure why this happens.

 

[hmmm27]

So that charges do get conducted away and stay on the metal balls. Insulators do not allow charges to flow through it.

Ah, okay thanks...

Now, from your original(sketched) diagram, you've got
(-+)(-+) [-]
Which makes it look like there's a capacitance between the individual balls ... that are metal-skinned ... and touching ... ... ... yes ?

[edit: seems to be discussed a bit earlier in the thread... not too sure of the OP's resolution]

 

[haruspex]

charges on each metal ball should equal in magnitude the charge on ebony rod.

I was suggesting that the charges on the balls would be significantly less than the charge on the rod. As the rod approaches, the polarisation increases gradually, so the induced charge magnitudes certainly won't equal the charge on the rod initially.
The question is, can we show that the net force on the nearer ball towards the rod exceeds the net force on the further ball towards the ball+rod combination?

 

[Delta2]

You can model the system of the two small metal sphere as an electric dipole with charges +q and -q separated initially by distance 2d where d the diameter of each sphere(because this charges are initially located at the two furthest edges of the two spheres, -q at the left edge of the left sphere and +q at the right edge of the right sphere).

Then you can use the condition that the total electric field in the center of this dipole is zero (coming from the condition that the total electric field in the interior of any conductor (viewing the two touching spheres as one conductor) should be zero in electrostatic equilibrium), to prove that +q<|Q|, Q being the (negative) charge of the rod.

P.S Of course in fact we don't have a dipole and the electric field is everywhere zero in between the two sphere (at their initial configuration when they are touching), not only at the center.

 

[vcsharp2003]

Ah, okay thanks...

Now, from your original(sketched) diagram, you've got
(-+)(-+) [-]
Which makes it look like there's a capacitance between the individual balls ... that are metal-skinned ... and touching ... ... ... yes ?

[edit: seems to be discussed a bit earlier in the thread... not too sure of the OP's resolution]

That part was clarified to me earlier in this thread. My solution diagram was not correct.

 

[vcsharp2003]

I was suggesting that the charges on the balls would be significantly less than the charge on the rod.

What is the reason for induced charge being less than charge on rod? Maybe the induced electric field causes charge redistribution to stop on the metal balls after only a small charge redistribution has occurred. After all, it's the electric field that will be the root cause of a charge to move.

The question is, can we show that the net force on the nearer ball towards the rod exceeds the net force on the further ball towards the ball+rod combination?

Of course, if induced charges are very small compared to charge on rod, then repulsion force on further ball due to -ve charges on rod will be higher than force of attraction on it from +ve charged metal ball. This would be due to Coulomb's law.

 

[vcsharp2003]

Then you can use the condition that the total electric field in the center of this dipole is zero (coming from the condition that the total electric field in the interior of any conductor (viewing the two touching spheres as one conductor) should be zero in electrostatic equilibrium), to prove that +q<|Q|, Q being the (negative) charge of the rod.

If we take a point in interior of two balls system then how would that prove +q<|Q|?

On the other hand, we can also say that as soon as a small amount of charges +q and -q have appeared on ends of two ball system then we will have a zero electric field inside the two ball system and hence further charge redistribution will stop. Free electrons inside this two ball system cannot move in a zero electric field space. Thus, +q<|Q|always.

 

[Delta2]

If we take a point in interior of two balls system then how would that prove +q<|Q|?

On the other hand, we can also say that as soon as a small amount of charges +q and -q have appeared on ends of two ball system then we will have a zero electric field inside the two ball system and hence further charge redistribution will stop. Free electrons inside this two ball system cannot move in a zero electric field space. Thus, +q<|Q|always.

For any point in the interior of the two balls, the electric field will be zero (while they still touching), I think we can still prove that q<|Q| ,but I guess the proof becomes more difficult. I believe a full treatment of this problem is worth of writing a paper that will appear at some journal of physics.

Yes it happens as you say, charge redistribution stops soon and then the balls start moving further from each other, the point is how to prove that the charge q will be smaller than Q, my attempt at post #17 is maybe a bit simplistic, but I think it gives a general idea of why q<|Q|.