Electrostatic force exerted on an electron inside a nucleus

[mangofan]

TL;DR Summary

Why is the distance from the center of the nucleus not taken into consideration?

This is SAMPLE PROBLEM 25-7 from "Physics" by Resnik, Halliday, and Krane, in the chapter "Electric Field and Coulomb's Law".

After describing the behavior of uniformly charged spherical shells:

A uniformly charged spherical shell exerts no electrostatic force on a point charge located anywhere inside the shell.

A uniformly charged spherical shell exerts an electrostatic force on a point charge outside the shell as if all the charge of the shell were concentrated in a point charge at its center.

follows a sample problem:

SAMPLE PROBLEM 25-7. The spherical nucleus of a certain atom contains a positive charge Ze in a volume of radius R.
Compare the force exerted on an electron inside the nucleus at radius 0.5R with the force at radius R for a nucleus in which (a) the charge density is constant throughout its volume, and (b) the charge density increases in direct proportion to the radius r.

The solution to (a) goes to say that the volume inside R/2 is 1/8 of the total volume, therefore the charge is 1/8 of the total charge and finally the ratio of the forces F(R/2)/F(R) is 1/8.

Now this doesn't make complete sense to me. The electrostatic force is also inversely proportional to the distance between the charges. Taking that into account, F(R/2)/F(R) = 1/8 * 4/1 = 1/2. Am I missing something? I would be surprised for the book to contain such an obvious error.

 

[TSny]

The solution to (a) goes to say that the volume inside R/2 is 1/8 of the total volume, therefore the charge is 1/8 of the total charge and finally the ratio of the forces F(R/2)/F(R) is 1/8.

Now this doesn't make complete sense to me. The electrostatic force is also inversely proportional to the distance between the charges. Taking that into account, F(R/2)/F(R) = 1/8 * 4/1 = 1/2. Am I missing something?

Your reasoning looks correct to me. To be picky, the force between two point charges is inversely proportional to the square of the distance of separation. But you took the square into account in your reasoning. So, your conclusion is correct.

I would be surprised for the book to contain such an obvious error.

It is a bit surprising. Are you using a recently published edition?

 

[kuruman]

It is a bit surprising. Are you using a recently published edition?

Haliday and Resnick is a venerable textbook that has been around since 1960. "Adequately proofread" would probably be more appropriate than "recently published" in this case.

 

[mangofan]

It's the 5th edition (the most recent) from 2002. Here is the complete text:

 

[kuruman]

Thank you for posting that.

Whoever wrote that, assumed that 1/8 of the charge is at distance R not R/2.
Whoever proofread that, did not catch the mistake.

 

[Mister T]

It's the 5th edition (the most recent) from 2002. Here is the complete text:

See if that same example problem is included in earlier editions.

Also, do you know which printing you have of the 5th edition?

 

[mangofan]

I've bought the book as new earlier this year and the pages look white, so it must have been printed quite recently. There's no info about a specific printing, the copyright section mentions 1960, 1962, ..., 1992, 2002.

Looking at the preface, it mentions that computing the force for a line of charge (volume of charge in the case above) was newly added in the 5th edition. There was no opportunity to fix the error.

 

[vanhees71]

You just have to solve for the electrostatic potential, which is very simple in these spherically symmetric examples. The potential is dependent only on $r$, where the center of the spherical charge distribution is taken to sit at the origin of the coordinate system. In spherical coordinates you thus simply have to solve
$$-\Delta \Phi(r)=-\frac{1}{r} [r \Phi(r)]''=\frac{1}{\epsilon}_0 \rho(r),$$
which can be done by two integrations. The constants of these integrations are determined by the condition that there is no singularity at $r=0$, if $\rho$ is continuous there, and that you can arbitrarily choose $\Phi(r) \rightarrow 0$ for $r \rightarrow \infty$.

 

[mangofan]

@vanhees71 , are you suggesting that more computation is needed and the answer for (1), constant charge density, is not $\frac{F(R/2)}{F(r)} = \frac{1}{2}$ as outlined above?

 

[vanhees71]

I don't know, what you are referring to here. The problems mentioned in the book are all of the form that you have a given spherically symmetric charge distribution $\rho(r)$ and want the electrostatic field everywhere. For that it's most easy to calculate the electrostatic potential, which by symmetry is a function of $r$ only (in spherical coordinates), i.e.,
$$-\frac{1}{r} [r \Phi(r)]''=\frac{1}{\epsilon_0} \rho(r).$$
The charge distribution is is assumed to vanish outside a sphere of radius $R$, i.e., $\rho(r)=0$ for $r>R$.

So for $r>R$ you get
$$[r \Phi(r)]''=0 \; \Rightarrow \; r \Phi(r)=C_1+C_2 r \; \Rightarrow \; \Phi(r)=\frac{C_1}{r} + C_2 \quad \text{for} \quad r>R.$$
We can choose an arbitrary additive constant anyway freely, and it's convenient to choose it such that $\Phi(r) \rightarrow 0$ for $r \rightarrow \infty$, i.e., we set $C_2=0$. For the electric field you get there
$$\vec{E}=-\vec{\nabla} \Phi=\frac{C_1}{r^2} \vec{e}_r.$$

In the charge-free space you have a Coulomb field, and Gauß's Law tells you that for any spherical shell with a radius $a>R$ you have
$$\int_{S(a)} \mathrm{d}^2 \vec{f} \cdot \vec{E}=4 \pi a^2 E_r(a)=4 \pi C_1=\frac{Q}{\epsilon_0} \; \Rightarrow \ \; C_1=\frac{Q}{4 \pi\epsilon_0}.$$
For $r<R$ you have to integrate the above equation including the given $\rho(r)$ twice, i.e.,
$$[r \Phi(r)]'=-\int_0^r \mathrm{d} r' r' \rho(r')/\epsilon_0 + C_1',$$
and then
$$r \Phi(r)=-\int_0^r \mathrm{d} r'' \int_0^{r''} \mathrm{d} r ' r' \rho(r')/\epsilon_0 + C_1' r + C_2'$$
or
$$\Phi(r)=-\frac{1}{r} \int_0^r \mathrm{d} r'' \int_0^{r''} \mathrm{d} r ' r' \rho(r')/\epsilon_0 + C_1'+\frac{C_2'}{r} \quad \text{for} \quad r<R.$$
Now there are no singularities at $r=0$, i.e., you must have $C_2'=0$. Further there's no surface charge at $r=R$ in addition to the "bulk" charge distribution $\rho(r)$ for $r<R$, and thus both $E_r$ and $\Phi$ are continuous at $r=R$, from which you get $C_1'$. For $\vec{E}=E_r \vec{e}_r$ you even don't need $C_1'$ at all.

Now take the first example,
$$\rho(r)=\rho_0=\frac{3 Q}{4 \pi R^3} \quad \text{f"ur} \quad r<R.$$
Then for $r<R$ you have
$$[r \Phi(r)]''=-\frac{\rho_0}{\epsilon_0} r \; \Rightarrow \; [r \Phi(r)]'=-\frac{\rho_0}{2 \epsilon_0} r^2 + C_1' \; \Rightarrow \; r \Phi(r)=-\frac{\rho_0}{6 \epsilon_0} r^3 + C_1' r.$$
Finally
$$\Phi(r)=-\frac{\rho_0}{6 \epsilon_0} r^2 + C_1'=-\frac{Q}{8 \pi \epsilon_0 R^3} r^2+C_1' \quad \text{for} \quad r<R.$$
For the electric field you thus get
$$E_r=-\Phi'(r)=\begin{cases} \frac{Q}{4 \pi \epsilon_0 R^3} r & \text{for} \quad r<R, \\ \frac{Q}{4 \pi \epsilon_0 r^2} & \text{for} \quad r \geq R, \end{cases}$$
and indeed $E_r$ is continuous at $r=R$, as already argued about above.

The 2nd example, where $\rho=C r$ for $r<R$ you can try to solve for yourself. It's a good exercise!

 

[TSny]

This sample problem is from chapter 25 of the text, which is the first chapter dealing with electricity and magnetism. The concepts of electric field and electric potential are introduced in later chapters. The tools available to solve the problem are Coulomb's law and the "shell theorem". The shell theorem is stated in the text (without derivation) a few paragraphs before the sample problem.

 

[mangofan]

Thanks @vanhees71 for the explanation and @TSny for the clarification.

Indeed, if we take the final formula where $E_r \propto r$, we get $\frac{F(R/2)}{F(R)} = \frac{1}{2}$, the same result as Coulomb's Law for the equivalent point charge.

 

[kuruman]

The posted wrong solution impies that, instead of using $\mathbf{F}=q\mathbf{E}$ for the Coulomb force on the charge, the expression $~\mathbf{F}=q \int_s \mathbf{E}\cdot\hat n~dA~$ was effectively used. Here is why. $$\begin{align}
&F(R)=q \int_{S_1} EdA=\frac{q}{\epsilon_0}\int_0^R~\rho(r)dV \nonumber \\
& F(R/2)=q \int_{S_2} EdA=\frac{q}{\epsilon_0}\int_0^{\frac{R}{2}}~\rho(r)dV \nonumber \\
&\implies \frac{F(R/2)}{F(R)}=\frac{\int_0^{\frac{R}{2}}~\rho(r)dV}{\int_0^R~\rho(r)dV}.\nonumber
\end{align}$$The expression erroneously gives the ratio of the forces as the ratio of enclosed charges.

 

[TSny]

The posted wrong solution impies that, instead of using $\mathbf{F}=q\mathbf{E}$ for the Coulomb force on the charge, the expression $~\mathbf{F}=q \int_s \mathbf{E}\cdot\hat n~dA~$ was used.

Maybe. But this seems strange to me. The incorrect expression $~\mathbf{F}=q \int_s \mathbf{E}\cdot\hat n~dA~$ is not even dimensionally consistent.

 

[Mister T]

I've bought the book as new earlier this year and the pages look white, so it must have been printed quite recently. There's no info about a specific printing, the copyright section mentions 1960, 1962, ..., 1992, 2002.

Looking at the preface, it mentions that computing the force for a line of charge (volume of charge in the case above) was newly added in the 5th edition. There was no opportunity to fix the error.

Look at the row of numbers at the bottom of the book's copyright page. The lowest number in the series is the number of the book printing. Each time a new printing is made known errors get corrected. You will sometimes find that even though you and your classmates and your professor all have the same edition, there will be slight differences due to different printings of the same edition.

 

[TSny]

I found a copy of the 5th edition in Spanish with a copyright date of 2002 and a printing date of 2007. It has the same error. So, the error has been around for quite a while.

 

[mangofan]

Here is the copyright page. @Mister T , are you referring to the "10" on the bottom of the page?

 

[Vanadium 50]

A common complaint is "Harumph! The publishers change editions too often!" Another common complaint is "Harumph! The errors never seem to get corrected!"



The solutions manual in particular has and has always had a number of mistakes.

 

[kuruman]

A common complaint is "Harumph! The publishers change editions too often!" Another common complaint is "Harumph! The errors never seem to get corrected!"



The solutions manual in particular has and has always had a number of mistakes.

Exactly! The errors never seem to be corrected because the publishers change editions too often.

 

[Mister T]

Here is the copyright page. @Mister T , are you referring to the "10" on the bottom of the page?

Yes. You have the 10th printing.