Basic electrostatic Force question

  • Thread starter leolaw
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  • #1
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Sphere A carries a net charge and sphere B is neutral. Both are conducting spheres and they are placed near each other on an insulated table. Which statement best describes the electrostatic force between them?

a) there is no force b/w them since one is neutral
b) there is a force of attraction
c) there is a force of repulsion
d) the force is attractive if A is charged positively and repulsive if A is charged negatively.

I think answer (a) is right because according to Coulomb's law, [tex] F = k\frac{q_1q_2}{r^2}[/tex], and since that B is neutral so its net charge should be 0, which makes the entire equation equals to zero.

But then i was thinking about attraction by induction between sphere A and B. And if it does happen, then answer (b) is also correct.

Which one of my idea is right?
 

Answers and Replies

  • #2
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Coulombs law, only accounts for point charges. Conducting spheres with evenly distributed charge on their surface can act like point charges at their center, but once the charge becomes non-uniform, this is no longer so.

Yes it is true that there is not net charge on the neutral sphere, but it is polarized by the charged sphere, thus creating an attraction.
 
  • #3
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So what happen to the other end of the sphere B? If one end is attracted to sphere A, then the other should have an equal and opposite force toward the other end, which makes the net force becomes zero right?

By the way, what exactly does polarization mean?
 
  • #4
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They don't cancell out. Couloumb force is propotional to [tex] \frac{1}{r^2}[/tex].

So the repulsives force is much less than the attractive force here.
 
  • #5
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good question. The exact definition of polarization can be found in your book, but essentially it is the effects of a slight shifting of charge within the molecules of a neutral insulator due to some electric field. I'm not sure if this term extends to conductors (I'll have to go look it up).

To give an example, I borrowed an illustration provided by Gokul43201 (Thanks Gokul!) from a previous thread on similiar topic, His illustration shows a similiar thing for plates instead of spheres (since plates are easier to draw)

Gokul43201 said:
Consider the 2 parallel plates (easier to draw), one of which (the one on the right) has an excess charge, +Q.

The second plate will experience charge separation, whereby a negative charge is induced on the side nearer the other plate, and a positive charge on the far side. This makes sense : opposite charges attract, etc.

Code:
     II            I                     
  ________      ________           
  |+    -|      |      |              
  |+    -|      |      |               
  |+ 0  -|      | +Q   |              
  |+    -|      |      |             
  |+    -|      |      |              
  --------      -------
since the negative charges are closer to plate 1 the attractive force is stronger then the repulsive force.
 

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