Electric potential of a negatively charged sphere

[bulbanos]

A negatively charged sphere is brought near an uncharged metal object. Positive charges appear on the side of the uncharged metal object nearest to the charged sphere, negative charges on the opposite side.
http://140.247.57.206/galileo/lgm/pi/CT_images/l5b.2.gif
I don't understand how the potential could be the same everywhere... There's a clear separation of charges which shoud produce a field imo.

 

[vincentchan]

first, let's make it clear.. the potential is not the same "EVERYWHERE", however, it has the same value INSIDE the metal. you can prove it even without mathematics.
In an zero resistant objecct, charge flows from high potential to low potential since nothing prevent it to do so.. so, if the potential is not same everywhere inside the metal, the charge will flow around, the high potential area will drop its potential because the charge is leaving.. the opposite is true for the low potential area.. until the system reaches an equilenvence point.. the charge will stop flowing around and the system has equipotential..
I think you confused the concept about electric field and potential.

 

[bulbanos]

first, let's make it clear.. the potential is not the same "EVERYWHERE", however, it has the same value INSIDE the metal. you can prove it even without mathematics.
In an zero resistant objecct, charge flows from high potential to low potential since nothing prevent it to do so.. so, if the potential is not same everywhere inside the metal, the charge will flow around, the high potential area will drop its potential because the charge is leaving.. the opposite is true for the low potential area.. until the system reaches an equilenvence point.. the charge will stop flowing around and the system has equipotential..
I think you confused the concept about electric field and potential.

it must be that :(
I see there some induced charge on the right side and the equivalent opposite charge on the other side. My idea is that there should be an electric field created due to that separation. And if you are an electron and you travel through that field, you are passing through a potential difference.

 

[cepheid]

Consider this rectangular slab of metal under the influence of an externally applied field E₀:

|-_______+|
|-_______+|
|-_______+|
|-_______+|
|-_______+|
|- <----E₁+|
|-_______+|
|-_______+|
|-_______+|
|-_______+|

----> E₀

Fig. 2.42

It makes sense that the externally applied field will induce movement in the charges within the conductor, creating a separation of charge. That separation of charge results in an electric field of its own, but the movement of charges continues only until this new field E₁ exactly balances the applied field E₀. Here's a more detailed explanation:

From Griffiths' Introduction to Electrodynamics, 3rd Ed:

(i) E = 0 inside a conductor. Why? Because if there were any field, those free charges would move, and it wouldn't be electrostaticsanymore. Well...that's hardly a satisfactory explanation; maybe all it proves is that you can't have electrostatics when conductors are present. We had better examine what happens when you put a conductor into an external electric field E₀ (Fig. 2.42). Initially, this will drive any free positive charges to the right, and negative ones to the left. (In practice, it is only the negative charges -- electrons -- that do the moving, but when they depart the right side is left with a net positive charge -- the stationary nuclei-- so it doesn't really matter which charges move; the effect is the same.) When they come to the edge of the material, the charges pile up: plus on the right side, minus on the left. Now, these induced charges will produce a field of their own, E₁, which, as you can see from the figure, is in the opposite direction to E₀. That's the crucial point, for it means that the field of the induced charge tends to cancel off the original field. Charge will continue to flow until this cancellation is complete, and the resultant field inside the conductor is precisely zero⁷. The whole process is practically instantaneous.

7 - Outside the conductor, the field is not zero, for here E₀ and E₁ do not cancel.

(ii) \rho = 0 inside a conductor. This follows from Gauss's law: \nabla \cdot \mathbf{E} = \rho/\epsilon_{0}. If E = 0, then so also is \rho. There is still charge around, but exactly as much plus charge as minus, so the net charge density in the interior is zero.

(iii) Any net charge resides on the surface. That's the only other place it can be.

(iv) A conductor is an equipotential. For if a and b are any two points within (or at the surface of) a given conductor, V(b ) - V(a) = -\int_a^b{\mathbf{E} \cdot d\mathbf{l}} = 0, and hence V(b ) = V(a).

I think that addresses exactly the question you were asking.

 

[bulbanos]

thank you!
so, from which I deduce that if the object was a dielectric it would have a different potential on the sides because the charges are not free to move. All they can do is shift a bit creating an electric field inside which does not cancel the field outside.

 

[cepheid]

Exactly...it partially cancels the field outside, acting like a poor conductor. In my first year physics book (not Griffiths), it says that dielectrics can become true conductors under *very* strong applied fields, and that this is called "dielectric breakdown" if I remember right.