Tangent Angles for Lines of Force, Two Equal Positive Charges

[Ben2]

Homework Statement

In Halliday & Resnick, \mathcal {Physics for Students of Science and Engineering}, In Fig 27-4, consider any two lines of force leaving the upper charge. If the angle between their tangents for points near the charge is \theta, it becomes \theta/2 at great distances. Verify this statement and explain it."

Relevant Equations

E = q/(4\(pi$\epsilon{0}$r^2)

For lines of force symmetric with respect to the angle bisector of "near" tangent lines: An adaptation of the figure suggests that a right triangle with hypotenuse parallel to the right-hand "far" tangent line is similar to the right triangle with hypotenuse parallel to the "near" tangent line, second side the "near" angle bisector and the third side horizontal or vertical as the case may be. By Euclid, this would establish the claim. To me it's not clear how this can be applied to the general case, or if calculus must be used. Thanks!

 

[kuruman]

Please post Fig 27-4 so that we know what you are talking about.

 

[haruspex]

https://www.academia.edu/49244478/Physics_Part_II
The figure is on p 668, the text on p 681.
E.g. consider a line that starts out almost straight down from the upper charge. It does appear that this will asymptotically head off horizontally. But I don’t see how to apply the hint in general.

 

[TSny]

I don't think the book's claim about $\theta$ becoming $\theta/2$ is correct. I think it's much more complicated than that. I believe you have to consider solid angles.

 

[kuruman]

Here is the figure. This has to do with Gauss's law. Enclose one charge with a gaussian sphere that has a radius much smaller than the charge separation. Repeat with a gaussian sphere of radius much larger than the charge separation. In the first case one essentially encloses a monopole $+Q$ and in the other a monopole $+2Q$.

(Edited after posting just the figure.)

 

[Ben2]

I've never been able to scan a document picture. Here haruspex draws the same conclusion as I do. TSny's comment is also helpful, in the sense H&R's problem editor seems to have made additional assumptions, at least on q. I smelled E as a function of x and y. But I've never used vectors in comparing solid angles. Thanks very much to all respondents!

 

[haruspex]

Here is the figure. This has to do with Gauss's law. Enclose one charge with a gaussian sphere that has a radius much smaller than the charge separation. Repeat with a gaussian sphere of radius much larger than the charge separation. In the first case one essentially encloses a monopole $+Q$ and in the other a monopole $+2Q$.

Ok, but let's see where that leads.
In each extreme, the field strength will be the same at all points around the sphere. So in each case there should be an equal number of field lines through each element of area.
The field lines emerging through the inner sphere between $\theta$ and $\theta+d\theta$ from the line joining the charges do so through a solid angle $2\pi\sin(\theta)d\theta$.
If those same lines reach the outer sphere between $\phi$ and $\phi+d\phi$ then, since the overall density has doubled, $\sin(\theta)d\theta=2\sin(\phi)d\phi$, leading to $\cos(\theta)=1+2\cos(\phi)$.

The textbook claim of a halved angle only works in 2D.

 

[TSny]

I think the problem is too difficult for the level of Halliday and Resnick’s text. Consider the following diagram

The two charges are labeled $q$. $r_1$ is the radius of an imaginary sphere centered on the top charge. We take $r_1$ small enough that the points on the sphere are "very near" the top charge. $r_2$ is the radius of a large sphere centered on the point between the two charges. $r_2$ is "very far" from the two charges.

We have a central electric field line running from the top charge vertically upward. It remains straight all the way to infinity. Imagine a "cone" of electric field lines all making an angle $\theta_1$ to the vertical field line. I've only drawn two of these lines, one tilted to the right and the other to the left. These field lines will be curved so that when they reach the spherical surface at $r_2$, they make an angle $\theta_2$ to the central field line. The book claims that $\theta_2$ will equal half $\theta_1$.

The cone of field lines intersects the two spherical surfaces to make the spherical caps shown in blue. Let $A_1$ and $A_2$ be the surface areas of these caps, respectively. If you've covered Gauss' law (in the next chapter of the text), then you should be able to show that the flux of electric field is the same through the two caps. From geometry, the areas of the spherical caps are $A_1 = 2 \pi r_1^2(1-\cos\theta_1)$ and $A_2 = 2 \pi r_2^2(1-\cos\theta_2)$.

Consider the field strength on each of the spherical surfaces and whether or not the field can be taken to be uniform over each surface. Equating the fluxes through the caps should lead to the correct relation between $\theta_1$ and $\theta_2$.

The result only applies to this case where we are making use of the vertical field line that is straight. If we chose two arbitrary field lines leaving the top charge so that both field lines are curved, the geometry is more complicated.

 

[bob012345]

If the angle between their tangents for points near the charge is $\theta$

This isn't defined. Tangent to what?

 

[haruspex]

If the angle between their tangents for points near the charge is $\theta$

This isn't defined. Tangent to what?

The tangents to the field lines through the two points, at the two points.

 

[bob012345]

Then assuming there are N lines around a single charge in the close case they are evenly spread around the sphere and in the far case N lines are spread evenly around a half sphere. In any plane cut through the sphere the angles between any two line's tangents should be half in the far case.

 

[haruspex]

Then assuming there are N lines around a single charge in the close case they are evenly spread around the sphere and in the far case N lines are spread evenly around a half sphere. In any plane cut through the sphere the angles between any two line's tangents should be half in the far case.

Evenly spread around a sphere means the same number per unit area. There's more area in a band around the equator than there is in a band subtending the same plane angle to the centre around a pole.
See posts #7 and #8.

 

[bob012345]

Evenly spread around a sphere means the same number per unit area. There's more area in a band around the equator than there is in a band subtending the same plane angle to the centre around a pole.
See posts #7 and #8.

True if this were a problem in Jackson but this is a HR problem which implies at far enough distance it looks just like a single charge of $2q$ and the lines are spread out evenly. Twice the lines over the sphere implies half the angle between any two lines over the single charge case. I would say it approaches exactly one half at infinity.

 

[haruspex]

True if this were a problem in Jackson but this is a HR problem which implies at far enough distance it looks just like a single charge of $2q$ and the lines are spread out evenly. Twice the lines over the sphere implies half the angle between any two lines over the single charge case.

But that isn't the question asked. The question is the angle that a given field line ends up at. Your approach does not address the marrying of individual field lines emanating from a charge to the field lines at great distance.

Edit: Skeptical, eh, @bob012345? So find the logical flaw in the method in posts #7 and #8.

 

[bob012345]

But that isn't the question asked. The question is the angle that a given field line ends up at. Your approach does not address the marrying of individual field lines emanating from a charge to the field lines at great distance.

Doesn't it say tangents between any two lines? Question #7 in chapter 27 of my copy of H&R asks why the lines of forces from fig. 27-4 when extended backwards, appear to radiate uniformly from the center of the figure.

 

[haruspex]

Doesn't it say tangents between any two lines?

If you know what each individual line does then you can deduce what the angle between any two lines in the same "vertical" plane does, and vice versa.

why the lines of forces from fig. 27-4 when extended backwards, appear to radiate uniformly from the center of the figure

They don't. It's the same error.
Note that the figure is a 2D projection. If lots of field lines were drawn, roughly evenly spaced around the spherical surface, they would not look evenly spaced in the projection: they would be denser near the equator. This applies to both the small sphere and the large one.

 

[bob012345]

If you know what each individual line does then you can deduce what the angle between any two lines in the same "vertical" plane does, and vice versa.

They don't. It's the same error.
Note that the figure is a 2D projection. If lots of field lines were drawn, roughly evenly spaced around the spherical surface, they would not look evenly spaced in the projection: they would be denser near the equator. This applies to both the small sphere and the large one, and such a diagram does not easily show how the one mesh maps to the other.

Then the error is in H&R which I'm skeptical of. That figure is a cut through one plane and it looks the same in any plane around that axis. If you pick any two lines of the top charge in that figure there is some distance where the angle is half between the tangents.

 

[haruspex]

Then the error is in H&R which I'm skeptical of.

H&R doesn't offer an actual proof, just handwaving, appealing to an inappropriate diagram. @TSny and I offer (effectively the same) proof. Don't just be sceptical, check the proof.

That figure is a cut through one plane and it looks the same in any plane around that axis.

But it does not correctly illustrate the field density distribution. It shows the same total field lines emanating between the equator and 5° latitude as between 85° and 90° latitude.

If you pick any two lines of the top charge in that figure there is some distance where the angle is half between the tangents.

We do not know how the figure was drawn. It might not be that accurate.
But supposing it is, how did you determine that?
If you pick two lines close together it may look like that because the derivative of the equation I gave in post #7 is $\sin(\theta)=2\sin(\phi)\frac{d\phi}{d\theta}$, which for small angle steps looks like $d\theta=2d\phi$. Try comparing where a line that emerges 20° below the horizontal ends up with where H&R predict, 35° above horizontal.

 

[bob012345]

But that isn't the question asked. The question is the angle that a given field line ends up at. Your approach does not address the marrying of individual field lines emanating from a charge to the field lines at great distance.

Edit: Skeptical, eh, @bob012345? So find the logical flaw in the method in posts #7 and #8.

I found the problem in my copy of H&R and it's completely different from the OP statement. I shall type it out. This is from the 1978 version of combines parts 1&2

In fig. 27-4 consider two neighboring lines of force leaving the upper charge at small angles with a straight line connecting the charges. If the angle between their tangents for points near the charge is $\theta$, it becomes $\frac{\theta}{\sqrt{2}}$ at great distances. Verify this statement and explain it. (Hint: Consider how the lines must behave both close to either charge and far from the charges.)

So, it's not one half. I'm wrong there. But do the methods of #7 and #8 lead to $\frac{\theta}{\sqrt{2}}$ and is that angle correct?

 

[kuruman]

Commensurate with the end-of-chapter questions in Halliday and Resnick, I think the expected argument is qualitative, hand waving and in 2d as is Fig. 27-4.

Construct Gaussian sphere of radius $R_1<<<d$ where $d$ is the separation between charges and enclose the top charge at the center. The electric field lines from the other charge can be ignored because they are much weaker than those form the enclosed charge. The electric field lines emerging from the surface of the sphere are by and large spherically symmetric. If they are not, make the radius smaller until they are. The electric field in any plane intersecting the sphere and containing the charge will have field lines as shown in the figure below left consistent with the idea that where field lines are closer together the field is stronger. As has been noted already, this is a 2d picture and we are looking at the in-plane electric field. The spherical symmetry of the point charge says that there is no "equator" where the lines will be closer together and no z-axis with respect to which spherical angle $\theta$ is defined.

Now construct another Gaussian sphere of radius $R_2>>>d$ (figure below right). Here we also have spherical symmetry but twice as much charge at the center and twice as many lines emerging from the surface. The red field lines from the second charge are interspersed with the black field lines from the first charge. The result is half the angle between field lines at large distances. I don't like this argument but it is what I think the book wants me to say.

 

[haruspex]

I found the problem in my copy of H&R and it's completely different from the OP statement. I shall type it out.

In fig. 27-4 consider two neighboring lines of force leaving the upper charge at small angles with a straight line connecting the charges. If the angle between their tangents for points near the charge is $\theta$, it becomes $\frac{\theta}{\sqrt{2}}$ at great distances. Verify this statement and explain it. (Hint: Consider how the lines must behave both close to either charge and far from the charges.) From the 1978 version of combines parts 1&2.

So, it's not one half. I'm wrong there. But do the methods of #7 and #8 lead to $\frac{\theta}{\sqrt{2}}$ and is that angle correct?

That version is worse, in that it is easily debunked. If it were true, the whole of the field lines from the upper charge, a complete $2\pi$, ends up compressed into an angle of $\sqrt 2\pi$, instead of just $\pi$. The field lines from the two charges would intersect.

See post #26.

 

[bob012345]

That version is worse, in that it is easily debunked. If it were true, the whole of the field lines from the upper charge, a complete $2\pi$, ends up compressed into an angle of $\sqrt 2\pi$, instead of just $\pi$. The field lines from the two charges would intersect.

But the problem is limited to a small angle near the pole, the center line connecting the charges, not any angle.

 

[bob012345]

It's not worth going further until we at least know the correct version of the problem statement.

 

[haruspex]

and in 2d

Quite. It is the correct result in a 2D space, not in our 3D one.

 

[bob012345]

What then is the correct far angle in the 3D universe? Not sure how to get it from the posts #7 and #8.

 

[haruspex]

But the problem is limited to a small angle near the pole, the center line connecting the charges, not any angle.

Ah, yes, I did not notice the two changes in the problem statement.
I would say they are two different problems.
The solution in posts #7 and #8 gives the same answer as H&R for the later problem.

 

[bob012345]

Ah, yes, I did not notice the two changes in the problem statement.
I would say they are two different problems.
The solution in posts #7 and #8 gives the same answer as H&R for the later problem.

Thanks. I'm still confused as to what you mean. Is that answer 1/2 and is that in 3D?

 

[haruspex]

Thanks. I'm still confused as to what you mean. Is that answer 1/2 and is that in 3D?

No, it's the answer H&R give for the revised problem, $\frac 1{\sqrt 2}$, in 3D.

 

[TSny]

A formula relating the two angles $\theta_1$ and $\theta_2$ in the figure of post #8 is $$\theta_2 = \cos^{-1}\left[ \frac{1}{2}(1+\cos\theta_1 )\right]$$
For example, a field line that leaves the upper charge horizontally ($\theta_1 = 90^o$) will end up with $\theta_2 = 60^o$.

If the angles are expressed in radians and if $\theta_1$ is small, then $\theta_2 \approx \large \frac{ \theta_1}{\sqrt{2}}$.

 

[Ben2]

Will study the whole thread, and thanks to all respondents! "Anticipating" later developments is common in math, but I'd not seen enough to realize it happens in physics.