Point charges placed inside a charged sphere

[A13235378]

Homework Statement

The figure represents a sphere of radius R uniformly charged with a positive charge. Inside there are two negative point charges (-Q each) placed on the same diameter of the sphere and equidistant from the center. The system is electrically neutral. Determine the distance x to which the negative charges from the center of the sphere must be for the system to be in electrostatic equilibrium.

Relevant Equations

Gauss Law
Eletric force

I traced a spherical X-ray Gaussian (green) where the negative charges were diametrically opposite. My question is this: I can transform the entire charge of the Gaussian sphere into a point charge placed in the center. So, can I analyze only the electrical forces of the two negative charges with the "new" point charge? Yes or No? And why?

 

[haruspex]

Wrt equilibrium, what can move? Are the two point charges at a fixed separation, but, otherwise, everything is just floating in space?

 

[kuruman]

Wrt equilibrium, what can move? Are the two point charges at a fixed separation, but, otherwise, everything is just floating in space?

I think both charges can move but only along a diameter. I imagine a very small hole drilled through a diameter of the sphere with the two charges inside it. One is looking for the equilibrium separation between the charges.

 

[haruspex]

I think both charges can move but only along a diameter. I imagine a very small hole drilled through a diameter of the sphere with the two charges inside it. One is looking for the equilibrium separation between the charges.

Yes, that's at least as reasonable a reading. But neither makes it a very interesting problem, or am I missing something? The sphere must be an insulator since we are told the charge is uniformly distributed.

 

[jbriggs444]

It is at least somewhat interesting. One cannot blindly treat the charged spherical volume as if it were a point charge of magnitude 2q.

I've discarded the idea that it is a uniformly charged spherical shell of zero thickness rather than a spherical volume. The equilibrium position for that case is not aesthetically pleasing. However, technically, "sphere" means the surface rather than the volume therein. So this could be the intended interpretation.

So, can I analyze only the electrical forces of the two negative charges with the "new" point charge? Yes or No? And why?

You cannot replace the two point charges with an equivalent charge at their midpoint. There is no such equivalency.

The shell theorem allows you to treat a uniformly charged shell (and, consequently, any collection of concentric uniformly charged shells) as if the charge were concentrated in the center. Two identical point charges by themselves do not comprise a uniformly charged shell.

You do have a uniformly charged spherical volume at hand. The equivalence would be valid for it. Be careful about the caveat for how the equivalency works for the effect on objects inside versus objects outside.

 

[kuruman]

In the mundane version problem one has three collinear charges, two are identical at the ends and one has opposite sign in the middle. The question is to find the distance between the end charges at which the net force on them is zero. If we assume that the situation is one-dimensional and the charges are constrained to be on the x-axis, then even though the net force on each charge may be zero, they are not in stable equilibrium. The electrostatic potential at one of the end charges must satisfy Laplace's equation in one dimension, $\dfrac{d\varphi^2}{dx^2}=0$. For stable equilibrium, the second derivative has to be positive, not zero. This problem asks us to find the position of stable equilibrium.

When the middle charge is replaced by a uniform spherical charge distribution, then Poisson's rather than Laplace's equation must be satisfied which makes it possible to have a second derivative greater than zero. This, I think, is what makes the problem interesting.

I found an answer by treating the potential in which one of the charges finds itself as the superposition of the potential inside the charged sphere and the potential due to the other charge: $\varphi(x)=\varphi_{\text{sphere}}+\varphi_q$ under the assumption that the two charges are symmetrically disposed about the center of the sphere, i.e. their separation is $2x$. Then I just minimized.

Note that in this interpretation it is possible to model the effect of the solid sphere as an effective point charge at the center. However the effective point charge must depend on the position of the actual charges, $$Q_{\text{eff}}(x)=\frac{x^2}{k}\left(−\frac{d\varphi_{\text{sphere}}}{dx}\right).$$One can also displace the charges symmetrically about the equilibrium position and find the period of small oscillations.

 

[TSny]

My question is this: I can transform the entire charge of the Gaussian sphere into a point charge placed in the center. So, can I analyze only the electrical forces of the two negative charges with the "new" point charge? Yes or No? And why?

I think you are asking whether or not you can replace the system by the three charges shown on the right in the picture below.

As helpers, we're not allowed to just give you the answer. From your studies can you come up with an argument for why this simplification to three point charges is valid or not valid?

 

[haruspex]

"sphere" means the surface rather than the volume therein.

Ha! Of course. But a small clarification:

The shell theorem allows you, for regions outside the shell, to treat a uniformly charged shell (and, consequently, any collection of concentric uniformly charged shells) as if the charge were concentrated in the center ...

 

[kuruman]

The equilibrium position for that case is not aesthetically pleasing.

Will you please clarify which is "that" case? Is it the uniformly charged shell of zero thickness or is it a uniformly charged sphere, meaning having charge 2Q uniformly distributed over its volume? Thanks.

 

[jbriggs444]

Will you please clarify which is "that" case? Is it the uniformly charged shell of zero thickness or is it a uniformly charged sphere, meaning having charge 2Q uniformly distributed over its volume? Thanks.

Uniformly charged shell of zero thickness is the one for which I dislike the resulting solution.

 

[kuruman]

Uniformly charged shell of zero thickness is the one for which I dislike the resulting solution.

That's what I thought. Is there even a possibility for an equilibrium solution in that case? The potential due to the shell is constant inside which means that the only electric field acting on one charge is due to the other charge.

 

[haruspex]

That's what I thought. Is there even a possibility for an equilibrium solution in that case? The potential due to the shell is constant inside which means that the only electric field acting on one charge is due to the other charge.

Only x=R.

 

[jbriggs444]

Only x=R.

Yes, that is what I had in mind. It is an uncomfortable solution because the electrostatic force there is undefined.

 

[Steve4Physics]

I’ll chip in. I believe the original question should be something like this:

A uniformly charged solid sphere [not a shell] of radius R has total charge +2Q.

Two point charges, each -Q, are [somehow] inserted so that they are on opposite sides of a diameter, each a distance x from the centre (x<R).

Find x such that the net force on each point charge is zero.

Based on this (without showing the few lines of working needed) I get x = R/2.

 

[kuruman]

Yes, that is what I had in mind. It is an uncomfortable solution because the electrostatic force there is undefined.

I am also uncomfortable with the discontinuity of the electric field. I am guessing that the author's intent was to have charge 2Q uniformly distributed over the volume of the sphere. I think what we have here is a helium atom description according to the plum pudding model.

It is reasonable to assume that in the early 1900's, when the model was espoused, the mathematics behind it were investigated. Quoting from the Wikipedia article, "In this model, the orbits of the electrons were stable because when an electron moved away from the centre of the positively charged sphere, it was subjected to a greater net positive inward force, because there was more positive charge inside its orbit (see Gauss's law). Electrons were free to rotate in rings which were further stabilized by interactions among the electrons, and spectroscopic measurements were meant to account for energy differences associated with different electron rings."

The dedication and persistence of these researchers is admirable. The anomalous Zeeman effect had already been observed but electron spin was about three decades in the future.

 

[hutchphd]

I'm slightly perplexed by the semantic argument here. The problem is stated correctly and unambiguously:

"The figure represents a sphere of radius R uniformly charged with a positive charge."

A sphere is a 3 dimensional object. "Uniformly charged" is unambiguous. If they wanted a spherical shell they would have so stated. What is the problem?

 

[jbriggs444]

Wiki says:

A sphere (from Greek σφαῖρα—sphaira, "globe, ball"[1]) is a geometrical object in three-dimensional space that is the surface of a ball (viz., analogous to the circular objects in two dimensions, where a "circle" circumscribes its "disk").

So there is at least some ambiguity about whether a spherical volume or a spherical shell is meant.

 

[hutchphd]

So a circle is a 1D object in your vernacular? (not trying to be argumentative but this seems wrong to me)

 

[hutchphd]

I see much to my surprise that the term is used in common usage for the ball or its surface. I am chagrinned by this lack of specificity but apologize for not being aware and casting aspersions. Somebody needs to fix this!

 

[jbriggs444]

So a circle is a 1D object in your vernacular? (not trying to be argumentative but this seems wrong to me)

Yes a circle is a 1d object.

A circle is the locus of points in a 2-d plane that are all at a particular distance from a central point. It does not include the interior.

[And yes, I know that usage varies. One can "cut a circle out of a piece of construction paper" and have a disc in one's hand. Or one can "draw a circle on this piece of graph paper" and have a thin line]

 

[hutchphd]

So should one say "the volume of a sphere" or "the volume in a sphere". Clearly this has me rattled.

 

[jbriggs444]

So should one say "the volume of a sphere" or "the volume in a sphere". Clearly this has me rattled.

Robustness principle: Be careful in what you send and generous in what you accept.

If you want to use "sphere" to refer to the volume, I have no major problem with that.

 

[kuruman]

I see much to my surprise that the term is used in common usage for the ball or its surface. I am chagrinned by this lack of specificity but apologize for not being aware and casting aspersions. Somebody needs to fix this!

I think that “someone” is the author of the problem especially in EM. One word specifying whether the distribution is of the “surface” or the “volume” variety will suffice.

 

[Gordianus]

Let me add confusion. When I was a teenager (eons ago) a sphere was a 3D object, not a surface. A circumference was the 1D object that enclosed a circle (2D). But this was customary when dinosaurs ruled the Earth.

 

[Steve4Physics]

Hi @A13235378 - if you're still reading this thread!

So, can I analyze only the electrical forces of the two negative charges with the "new" point charge? Yes or No? And why?

Yes, by applying the shell theorem (read-up if needed).

You can then balance [the repulsion on one of the negative charges by the other] against [the attraction of that negative charge by your "new" point charge].

 

[kuruman]

Let me add confusion. When I was a teenager (eons ago) a sphere was a 3D object, not a surface. A circumference was the 1D object that enclosed a circle (2D). But this was customary when dinosaurs ruled the Earth.

In what sense is the circumference of a circle a 1D object if a point on it requires two independent coordinates to be specified?

That’s a definition for people who are accustomed to going around in circles.

 

[haruspex]

In what sense is the circumference of a circle a 1D object if a point on it requires two independent coordinates to be specified?

It depends whether you consider a circle as a topological entity (a 1D manifold) or its Euclidean embodiment. Topologists speak of "round circles".

 

[haruspex]

I see much to my surprise that the term is used in common usage for the ball or its surface. I am chagrinned by this lack of specificity but apologize for not being aware and casting aspersions. Somebody needs to fix this!

It doesn’t help that in this question we are told there are point charges in the sphere which, from the reference to equilibrium, one might think are able to move. Yet there is no mention of a narrow hole drilled through it.
In hindsight, it doesn’t actually say they can move; it only asks for electrostatic equilibrium.

 

[kuruman]

It doesn’t help that in this question we are told there are point charges in the sphere which, from the reference to equilibrium, one might think are able to move. Yet there is no mention of a narrow hole drilled through it.
In hindsight, it doesn’t actually say they can move; it only asks for electrostatic equilibrium.

I agree, there is no mention of a hole or of a wire on which two charged beads can freely slide. I conjured up the hole as a constraint to convert this to a one-dimensional problem. No generality is lost because the only direction that counts is the diameter on which the charges are situated. I assume that by electrostatic equilibrium is meant stable electrostatic equilibrium. This means that a small displacement in a given direction results in a restoring force (or torque). There is no restoring force if an electron is displaced on the surface of an equipotential shell at which the radial force is zero.

I outlined my method in #6. Do you see any glaring infelicities?

 

[haruspex]

I assume that by electrostatic equilibrium is meant stable electrostatic equilibrium.

I see no need for such an assumption. It just means there's no net electrostatic force on each particle.