Chapter 2 Griffiths EM Problem: E-Field from a charged ring

[warhammer]

Homework Statement

Problem 2.5 from Griffiths EM:

Find Electric Field at a distance z above center of s circular loop of radius r that carries uniform line charge λ

Relevant Equations

E= $frac{1/4πϵ} * ( \int {1/r^2} ) \hat r \ dq$

Hi.

I have solved the problem below as shown in the attached image. However I'm at a loss to figure out where I am making a mistake, and I know it is indeed a big goof up. Requesting guidance over identification and rectification of this big goof up.

(Edit- I can solve this problem in the component form by formulating that only vertical components remain while all horizontal ones are canceled by each other. My aim is to solve this problem in the generic $\vec r$ & $\hat r$ style which is used by Griffiths in his Examples for Chapter 2 to achieve the same result wrt vertical components)

 

[Charles Link]

Suggest you start over. With what you have, you need to so an integral around the ring with angle $\phi$, rather than $\theta$. This one is not a difficult one to simply write out the answer, taking advantage of the symmetry, so that the result will point in the z direction.

 

[kuruman]

Yes, your mistake was in confusing the angles. Note that angle $\theta$ is constant for any element $dq$ around the ring. You made an early error in writing this charge element. It should be $dq=\lambda~r~d\phi.$

 

[Orodruin]

Also, it needs to be pointed out that $\hat r$ is not a constant vector. This needs to be kept in mind when doing integrals in cylinder coordinates.

 

[warhammer]

Homework Statement:: Problem 2.5 from Griffiths EM:

Find Electric Field at a distance z above center of s circular loop of radius r that carries uniform line charge λ
Relevant Equations:: E= $frac{1/4πϵ} * ( \int {1/r^2} ) \hat r \ dq$

Hi.

I have solved the problem below as shown in the attached image. However I'm at a loss to figure out where I am making a mistake, and I know it is indeed a big goof up. Requesting guidance over identification and rectification of this big goof up.

(Edit- I can solve this problem in the component form by formulating that only vertical components remain while all horizontal ones are canceled by each other. My aim is to solve this problem in the generic $\vec r$ & $\hat r$ style which is used by Griffiths in his Examples for Chapter 2 to achieve the same result wrt vertical components)

@Charles Link & @kuruman Thank you so much for your prompt responses. I have a couple of questions just to ensure I have understood what you both have advised correctly:

i) I should have used Φ instead of θ as the former is a angle on the plane of circle that actually sweeps around from 0--> 2π to give the circle, whereas the latter is a fixed angle. I cannot use θ also because it is not in the circle's plane that is sweeping around. Is this correct?

ii) Using the corrected approach, how will then the second integral comprising of $\int frac { (r * \hat r )/ cursive r} \ dΦ$ turn out to zero mathematically?

 

[Orodruin]

ii) Using the corrected approach, how will then the second integral comprising of ∫frac(r∗r^)/cursiver dΦ turn out to zero mathematically?

See post #4.

Edit: The correct LaTeX for what you want to write is
\int \frac{r \hat r}{\mathscr r} d\phi
$$
\int \frac{r \hat r}{\mathscr r} d\phi
$$

 

[warhammer]

See post #4.

Edit: The correct LaTeX for what you want to write is
\int \frac{r \hat r}{\mathscr r} d\phi
$$
\int \frac{r \hat r}{\mathscr r} d\phi
$$

Thank you for your response. Based on your thoughts, $\hat r$ is not constant, in the sense that its direction is not fixed and as we go along the circle its direction changes so to speak. So along the length of the circle 2 symmetric elements point in opposite direction (and by symmetry) which cancels them all out.

I hope my interpretation of your response is correct..

 

[Orodruin]

Thank you for your response. Based on your thoughts, $\hat r$ is not constant, in the sense that its direction is not fixed and as we go along the circle its direction changes so to speak. So along the length of the circle 2 symmetric elements point in opposite direction (and by symmetry) which cancels them all out.

I hope my interpretation of your response is correct..

Yes. Since the directions of the basis vectors in curvilinear coordinates change, you cannot treat them as constants and bring them out of sn integral as you can do with Cartesian basis vectors. In fact, it is not even clear what $\hat r$ would mean outside the integral as the point where it would be considered would be ambiguous.

Edit: What you can do is to express the curvilinear basis in a Cartesian basis and then take the Cartesian basis vectors out of the integral as they are constant.

 

[warhammer]

Yes. Since the directions of the basis vectors in curvilinear coordinates change, you cannot treat them as constants and bring them out of sn integral as you can do with Cartesian basis vectors. In fact, it is not even clear what $\hat r$ would mean outside the integral as the point where it would be considered would be ambiguous.

Edit: What you can do is to express the curvilinear basis in a Cartesian basis and then take the Cartesian basis vectors out of the integral as they are constant.

I see it now. So for fun sake were I to also prove this symmetric reasoning in terms of mathematics I should drop usage of $\hat r$ and instead use our Cartesian System..

So I think if I do that, my $\hat r$ should be $= z* \hat z - (x* \hat x + y* \hat y)$ as the circle technically lies within the xy plane?

 

[Orodruin]

I see it now. So for fun sake were I to also prove this symmetric reasoning in terms of mathematics I should drop usage of $\hat r$ and instead use our Cartesian System..

So I think if I do that, my $\hat r$ should be $= z* \hat z - (x* \hat x + y* \hat y)$ as the circle technically lies within the xy plane?

Note: You can still use cylinder coordinates to perform the integral.

 

[kuruman]

Alternatively, you can find the electric field through the potential. You can immediately write $$V(z)=\frac{1}{4\pi\epsilon_0}\int_0^{2\pi}\frac{\lambda~R~d\phi}{\sqrt{R^2+z^2}}=\frac{1}{4\pi\epsilon_0}\frac{2\pi R~\lambda}{\sqrt{R^2+z^2}}.$$The desired electric field is the negative derivative of the potential with respect to $z$.

In this symmetric case it doesn't really matter which way you do it because you have to do one integral either way. However, if you have to do 3 separate integrals to find the 3 components of the electric field, it might be easier to do 1 integral to find the potential and then take 3 derivatives for the components.

 

[warhammer]

Note: You can still use cylinder coordinates to perform the integral.

No that I got pretty well, I meant to ask if what was stated in #9 was fine or not to evaluate the integral keeping my basis vectors constant

 

[warhammer]

Alternatively, you can find the electric field through the potential. You can immediately write $$V(z)=\frac{1}{4\pi\epsilon_0}\int_0^{2\pi}\frac{\lambda~R~d\phi}{\sqrt{R^2+z^2}}=\frac{1}{4\pi\epsilon_0}\frac{2\pi R~\lambda}{\sqrt{R^2+z^2}}.$$The desired electric field is the negative derivative of the potential with respect to $z$.

In this symmetric case it doesn't really matter which way you do it because you have to do one integral either way. However, if you have to do 3 separate integrals to find the 3 components of the electric field, it might be easier to do 1 integral to find the potential and then take 3 derivatives for the components.

Indeed, using the Potential to find the Field is a much cleaner way when one uses alternative approach. However if you do not mind me asking a silly question, why was only $V(z)$ considered and not $V(x)$ or $V(y)$

 

[kuruman]

So I think if I do that, my $\hat r$ should be $= z* \hat z - (x* \hat x + y* \hat y)$ as the circle technically lies within the xy plane?

It should be $\mathbf{\hat r}=\cos\phi ~ \mathbf{\hat x}+\sin\phi~\mathbf{\hat y}$, where $\phi$ is measured from the $x$-axis in the xy-plane.

Indeed, using the Potential to find the Field is a much cleaner way when one uses alternative approach. However if you do not mind me asking a silly question, why was only $V(z)$ considered and not $V(x)$ or $V(y)$

Because we are asked to find the field on the z-axis where x and y are zero. Therefore, the potential can only be a function of z on the axis.

Edit: For off-axis points, one would write V(x,y,z) or in cylindrical coordinates V(r,z). Because of the azimuthal symmetry of the charge distribution, there would be no dependence on φ.

 

[hutchphd]

t should be r^=cos⁡ϕx^+sin⁡ϕ y^, where ϕ is measured from the x-axis in the xy-plane.

I disagree. By the OP definition r has a $\hat z$ component and that is the only surviving term, giving the result. Each of you has the Physics correct.

 

[warhammer]

It should be $\mathbf{\hat r}=\cos\phi ~ \mathbf{\hat x}+\sin\phi~\mathbf{\hat y}$, where $\phi$ is measured from the $x$-axis in the xy-plane.

But shouldn't the separation vector $\mathscr {\hat r}$ be what I mentioned in #9.. I'm imagining it as a point moving from source point present on the circle in xy plane to destination point on the z axis. Is this visualisation wrong?

(Edit- I think in the original post and different places I may have muddled up $\hat r$ & $\mathscr {\hat r}$ )

 

[Orodruin]

Alternatively, you can find the electric field through the potential. You can immediately write $$V(z)=\frac{1}{4\pi\epsilon_0}\int_0^{2\pi}\frac{\lambda~R~d\phi}{\sqrt{R^2+z^2}}=\frac{1}{4\pi\epsilon_0}\frac{2\pi R~\lambda}{\sqrt{R^2+z^2}}.$$The desired electric field is the negative derivative of the potential with respect to $z$.

In this symmetric case it doesn't really matter which way you do it because you have to do one integral either way. However, if you have to do 3 separate integrals to find the 3 components of the electric field, it might be easier to do 1 integral to find the potential and then take 3 derivatives for the components.

I am a big fan of potentials, but this does need a bit of a caveat to go along. As it is evaluating the potential along the z-axis only, it will only be able to give you the z-component as the other partial derivatives will remain unknown. Of course, you can argue that all other components are zero by symmetry (arguably the best type of argument), but that argument does need to be added during the solution.

This is precisely what got the OP confused here:

Indeed, using the Potential to find the Field is a much cleaner way when one uses alternative approach. However if you do not mind me asking a silly question, why was only $V(z)$ considered and not $V(x)$ or $V(y)$

This is a good observation.

 

[Orodruin]

Edit: For off-axis points, one would write V(x,y,z) or in cylindrical coordinates V(r,z). Because of the azimuthal symmetry of the charge distribution, there would be no dependence on φ.

The point is that to find the field components in the off-axis direction, you will need to know how the potential varies as you go off-axis (albeit sufficient to know the linear dependence near the axis). With the potential you computed it is not enough to find the components of the field perpendicular to the axis without additional symmetry arguments.

 

[kuruman]

The point is that to find the field components in the off-axis direction, you will need to know how the potential varies as you go off-axis (albeit sufficient to know the linear dependence near the axis). With the potential you computed it is not enough to find the components of the field perpendicular to the axis without additional symmetry arguments.

I agree.

 

[kuruman]

I disagree. By the OP definition r has a $\hat z$ component and that is the only surviving term, giving the result. Each of you has the Physics correct.

Yes, I don't particularly like Griffiths's "curly r" notation so I tend to ignore it when it raises its head. I stand corrected.

 

[vela]

So I think if I do that, my $\hat r$ should be $= z* \hat z - (x* \hat x + y* \hat y)$ as the circle technically lies within the xy plane?

I'm not sure if you meant to have the hat on the (script) r there. Using Griffith's notation, you would have
\begin{align*}
\mathbf r &= z\,\mathbf{\hat z} \\
\mathbf r' &= r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf {\hat y} \\
\mathbf d &= \mathbf r - \mathbf r' = z\,\mathbf {\hat z} - (r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf{\hat y}) \\
\hat{\mathbf d} &= \mathbf d/\| \mathbf d \| = \frac{z\,\mathbf {\hat z} - (r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf{\hat y})}{\sqrt{r^2+z^2}} \\
\end{align*} where I used $\mathbf d$ to represent Griffith's script-r character. It's just the direction $\hat{\mathbf d}$ which appears in the integral.

 

[Orodruin]

I'm not sure if you meant to have the hat on the (script) r there. Using Griffith's notation, you would have
\begin{align*}
\mathbf r &= z\,\mathbf{\hat z} \\
\mathbf r' &= r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf {\hat y} \\
\mathbf d &= \mathbf r - \mathbf r' = z\,\mathbf {\hat z} - (r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf{\hat y}) \\
\hat{\mathbf d} &= \mathbf d/\| \mathbf d \| = \frac{z\,\mathbf {\hat z} - (r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf{\hat y})}{r^2+z^2} \\
\end{align*} where I used $\mathbf d$ to represent Griffith's script-r character. It's just the direction $\hat{\mathbf d}$ which appears in the integral.

Missing a square root in the denominator in the last expression I believe.

 

[vela]

Yes, I did. I fixed it.

 

[kuruman]

$$\begin{align}\mathbf r &= z\,\mathbf{\hat z} \nonumber \\
\mathbf r' &= r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf {\hat y}\nonumber \end{align}$$

I don't kow how to process this. The first equation tells me that the magnitude of $\mathbf{r}$ is $z$, i.e. $r=z$. If I put that in the second equation, I get
$$\mathbf r'= z \cos\phi \,\mathbf {\hat x} + z \sin\phi\,\mathbf {\hat y}$$ which is wrong. That is why I am not comfortable with the Griffiths notation.

 

[vela]

Yeah, I noticed in this particular problem Griffiths used $r$ to denote the radius of the ring, so that's what I used here.

 

[warhammer]

The point is that to find the field components in the off-axis direction, you will need to know how the potential varies as you go off-axis (albeit sufficient to know the linear dependence near the axis). With the potential you computed it is not enough to find the components of the field perpendicular to the axis without additional symmetry arguments.

So what I infer from your response here is that computation of field is a much better option than Potential since the former delivers more information as compared to the latter or did you mean something else?

 

[warhammer]

I'm not sure if you meant to have the hat on the (script) r there. Using Griffith's notation, you would have
\begin{align*}
\mathbf r &= z\,\mathbf{\hat z} \\
\mathbf r' &= r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf {\hat y} \\
\mathbf d &= \mathbf r - \mathbf r' = z\,\mathbf {\hat z} - (r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf{\hat y}) \\
\hat{\mathbf d} &= \mathbf d/\| \mathbf d \| = \frac{z\,\mathbf {\hat z} - (r \cos\phi \,\mathbf {\hat x} + r \sin\phi\,\mathbf{\hat y})}{\sqrt{r^2+z^2}} \\
\end{align*} where I used $\mathbf d$ to represent Griffith's script-r character. It's just the direction $\hat{\mathbf d}$ which appears in the integral.

Yes. As you can see #16, I think I muddied it up with curly 'r' and added the hat over there. To expound on the same I think this should be fine:
Curly $\vec r$ $=z* \hat z$ $(- x* \hat x + y* \hat y)$

Curly $\hat r$ $=$frac({curly \vec r}/{√(z^2 + (x+y)^2)##.

Edit- I think due to confusions that arise on curly r usage, the curly r is merely a separation vector no? Whereas the \hat r that you have described above is a basis vector thing. Griffiths goes like curly \hat r=frac{\vec r} {r}

Edit 2- @vela I think I have another confusion to request guidance on. You have used cylindrical system in your response and then expressed it in cartesian basis vectors through the transformations. What I have based my response on is the assertion that the circle is contained within xy plane with center as origin and z axis perpendicular through it; the source element "on" circle represented as $x* \hat x + y* \hat y$ (because the circle too is on the xy plane and it must have some locative abscissa and ordinate). Then the destination is a point on the z axis which is represented by $z* \hat z$. Using this destination and source terminology I based my "curly r" vector above. Is this approach incorrect, and if yes then if you could please also guide how and why...

 

[warhammer]

I don't kow how to process this. The first equation tells me that the magnitude of $\mathbf{r}$ is $z$, i.e. $r=z$. If I put that in the second equation, I get
$$\mathbf r'= z \cos\phi \,\mathbf {\hat x} + z \sin\phi\,\mathbf {\hat y}$$ which is wrong. That is why I am not comfortable with the Griffiths notation.

To be honest there are moments where even I feel confused with this approach, mostly because one essentially uses the same letters for representing 2 distinct quantities and this approach sometimes lengthens a problem greatly.

If it is appropriate for me to ask here, what would be your suggestions for me based on your own experiences on using the book for strengthening EM concepts..

 

[kuruman]

If it is appropriate for me to ask here, what would be your suggestions for me based on your own experiences on using the book for strengthening EM concepts..

It is appropriate to ask here but the answer you will get from me is that Griffiths's "curly r" is the main reason I did not adopt the book when I taught intermediate E&M. I like his style and I loved teaching Quantum Mechanics out of his QM book, but I just couldn't bring myself to go along the "curly r" in the E&M book.

My suggestion would be to use the commonplace notation where unprimed coordinates are used for field points and primed coordinates for source points. Superposition of contributions from spatially distributed sources is done by integrating over primed coordinates. For example, the contribution to the electric field from charge element $dq$ located at point $mathbf{r'}$ to field point P located at $\mathbf{r}$ would be $$d\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{dq~(\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^3}.$$In this particular problem, the relevant vectors are in the Cartesian representation $$\begin{align} & \mathbf{r}=z~\mathbf{\hat z} \nonumber \\ & \mathbf{r'}=R\cos\phi'~\mathbf{\hat x}+R\sin\phi' ~\mathbf{\hat y}.\nonumber \end{align}$$Then $$d\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{\lambda R~d\phi'~(-R\cos\phi'~\mathbf{\hat x}-R\sin\phi' ~\mathbf{\hat y}+ z~\mathbf{\hat z})}{\left(R^2+z^2\right)^{3/2}}.$$ When you integrate over $\phi'$ from zero to $2\pi$, the first two terms disappear and all that is left is the z-component. This is the formal way of showing that the electric field has only a z-component on the z-axis.

 

[warhammer]

It is appropriate to ask here but the answer you will get from me is that Griffiths's "curly r" is the main reason I did not adopt the book when I taught intermediate E&M. I like his style and I loved teaching Quantum Mechanics out of his QM book, but I just couldn't bring myself to go along the "curly r" in the E&M book.

My suggestion would be to use the commonplace notation where unprimed coordinates are used for field points and primed coordinates for source points. Superposition of contributions from spatially distributed sources is done by integrating over primed coordinates. For example, the contribution to the electric field from charge element $dq$ located at point $mathbf{r'}$ to field point P located at $\mathbf{r}$ would be $$d\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{dq~(\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^3}.

Thank you so much for your response. I resonate greatly with your thoughts here. I think this is a more clutter free approach as well!

$$In this particular problem, the relevant vectors are in the Cartesian representation $$\begin{align} & \mathbf{r}=z~\mathbf{\hat z} \nonumber \\ & \mathbf{r'}=R\cos\phi'~\mathbf{\hat x}+R\sin\phi' ~\mathbf{\hat y}.\nonumber \end{align}$$Then $$d\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{\lambda R~d\phi'~(-R\cos\phi'~\mathbf{\hat x}-R\sin\phi' ~\mathbf{\hat y}+ z~\mathbf{\hat z})}{\left(R^2+z^2\right)^{3/2}}.$$ When you integrate over $\phi'$ from zero to $2\pi$, the first two terms disappear and all that is left is the z-component. This is the formal way of showing that the electric field has only a z-component on the z-axis.

This I believe satisfies what I set out to achieve, a brutely mathematical way to show the cancellation rather than symmetry usage blindly. A follow up to this, then is my visualisation in #16 wrong? Or is it a case of mine being in "curly r" with Cartesian coordinates and yours in regular r indicative of cylindrical system? (I'm sorry but still trying to grasp this more intuitively hence the multiple confirmations & persistence from my side)

 

[Orodruin]

So what I infer from your response here is that computation of field is a much better option than Potential since the former delivers more information as compared to the latter or did you mean something else?

No, that was not the point. The potential is many times easier to compute. The point made was that you cannot compute the potential along a single line if you want to know what all field components are.

 

[Orodruin]

rather than symmetry usage blindly

There is nothing blind about using symmetry arguments. In fact, I would claim that using symmetry arguments is at least as fundamental to a physicist’s toolbox as being able to do particular integrals. Probably even more important.

 

[warhammer]

There is nothing blind about using symmetry arguments. In fact, I would claim that using symmetry arguments is at least as fundamental to a physicist’s toolbox as being able to do particular integrals. Probably even more important.

I think of late I have developed a fanatical urge to follow a robustly mathematical approach in such questions where stuff like cancellations etc occur mathematically and not sleight of hand (which is a misconception/mental block on my part).

I think this habit is extremely unhealthy for someone who wishes to keep continuing with Physics, because Physics is not Mathematics even though the latter is the former's language. This is also a stumbling issue when I'm solving other problems, because I run into trouble with this apparent "rigidly mathematical" approach as per my retarded thoughtprocess instead of using symmetry as an effective tool.. Just thought of sharing this with you and others in a more detailed way here...

 

[Orodruin]

I think of late I have developed a fanatical urge to follow a robustly mathematical approach in such questions where stuff like cancellations etc occur mathematically and not sleight of hand (which is a misconception/mental block on my part).

Symmetry arguments have a robust mathematical foundation in group theory. There is nothing sleight of hand about it.

 

[warhammer]

Symmetry arguments have a robust mathematical foundation in group theory. There is nothing sleight of hand about it.

Indeed, which is why I referred to it as :) -

(which is a misconception/mental block on my part)