Electrostatics- Charged Identical Spheres

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When two identical conductive spheres are briefly connected, they share their charge equally due to their symmetry and identical surface areas. This process occurs because the electric potential across both spheres equalizes, resulting in a charge distribution of q/2 on each sphere if one initially has charge q and the other is neutral. The charges move to the surface of the spheres, which is a characteristic of conductors, ensuring that the total charge is evenly distributed. Understanding this requires knowledge of electric potential and the behavior of charges in conductors, which can be explored through integration of electric potential. Overall, the equal sharing of charge is a fundamental principle in electrostatics.
Isaac Sanctis
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In electrostatics, why is it that identical spheres that are touched for a brief moment or connected by a conducting wire (wire not gaining any charge) should have an equal charge? Why is that they come in an equilibrium of charge?
For example - If we take 2 identical spheres one bearing a charge q and the other neutral. The spheres are touched for a brief moment after which they are separated. Then it is stated that they both bear a charge of q/2 because they are identical ? But why so ?
 
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If they were not, the resulting electric field would induce a current which equalises the charges. If the spheres are reasonably conductive, this process happens very fast.
 
Could you please elaborate?
 
Which part do you have problems understanding?
 
All what you said... This topic has just been started with me in my classes and as I went through the Fundamentals of Physics (6th edition) Electric Charge, it's not stated why two identical spheres should have a equally shared net charge. And I have no knowledge of electric fields whatsoever.
 
Isaac Sanctis said:
All what you said... This topic has just been started with me in my classes and as I went through the Fundamentals of Physics (6th edition) Electric Charge, it's not stated why two identical spheres should have a equally shared net charge. And I have no knowledge of electric fields whatsoever.
this requires calculation. EM has many integration. First, you have to prove that there is no charge not on the surface of a conductor. I think this is proved by integrating electric potential of a sphere. When electric potential is minimum along an axis, it is maximum on another axis (not sure about sphere, I only calculated the ring). This solution is very important. It explains why charges in conductor must be on surface.

So, back to your question, I think you can think like this, the electric potential will push the charges on to the surface of the conductor. As you touch two spheres together, they are combined together, become one conductor, all charges go on surface evenly, when you separate them, charges on each sphere is q/2, reason is the surface area of two spheres are the same, also sphere is perfectly symmetric, that makes charges evenly distributed when two spheres are in contact.

hope you understand, you can simply integrate electric potential of a point lying on an axis, then use find out the maxima. then do this integration again for a point on another axis, then find out maxima again. For both cases, they should give different formula, and there should be at least one of the two solutions is not minimum.
 
sunmaggot said:
This solution is very important. It explains why charges in conductor must be on surface.
Can you elaborate on this point please ?
Thank you sunmaggot.
Apart from that I did understand everything except the below
"I think you can think like this, the electric potential will push the charges on to the surface of the conductor."

and

"hope you understand, you can simply integrate electric potential of a point lying on an axis, then use find out the maxima. then do this integration again for a point on another axis, then find out maxima again. For both cases, they should give different formula, and there should be at least one of the two solutions is not minimum."

Also I have no idea on electric potential but just given a formula i.e. kq/r
 
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