Find the electric field at a point in 3 dimensional space

[kuruman]

... but can you check the integration and point out where its wrong..?

I can, but only if you set up the integrals referring to a decent diagram which I have provided below. This, as most of the diagrams I create to post on PF, are made with PowerPoint.

 

[tellmesomething]

I can, but only if you set up the integrals referring to a decent diagram which I have provided below. This, as most of the diagrams I create to post on PF, are made with PowerPoint.


View attachment 345309

Following this
$$ R√2 tan \theta = x $$

$$ \frac {dx} {d\theta} = R√2 sec²\theta $$

$$ r= \frac { R√2} {cos \theta} $$

$$ dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 } $$

$$ \int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2} $$

$$ E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$

$$ E_{hori}= \frac {\lambda cos \phi } { 4π\epsilon R √2} $$

$$ E_{vert} =\frac { -\lambda } {4 π \epsilon R√6} (√3 + 1) $$

$$ E_{hori} =\frac {\lambda} { 4 π \epsilon R√3} $$

 

[tellmesomething]

Following this
$$ R√2 tan \theta = x $$
$$ \frac {dx} {d\theta} = R√2 sec²\theta $$
$$ r= \frac { R√2} {cos \theta} $$
$$ dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 } $$
$$ \int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2} $$
$$ E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$
$$ E_{hori}= \frac {\lambda cos \phi } { 4π\epsilon R √2} $$
$$ E_{vert} =\frac { -\lambda } {4 π \epsilon R√6} (√3 + 1) $$
$$ E_{hori} =\frac {\lambda} { 4 π \epsilon R√3} $$

Please let me know if you're seeing this executed ....im only seeing the tex commands at my end

 

[tellmesomething]

Following this
$$ R√2 tan \theta = x $$

$$ \frac {dx} {d\theta} = R√2 sec²\theta $$

$$ r= \frac { R√2} {cos \theta} $$

$$ dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 } $$

$$ \int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2} $$

$$ E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$

$$ E_{hori}= \frac {\lambda cos \phi } { 4π\epsilon R √2} $$

$$ E_{vert} =\frac { -\lambda } {4 π \epsilon R√6} (√3 + 1) $$

$$ E_{hori} =\frac {\lambda} { 4 π \epsilon R√3} $$

This is what it should be showing incase it didnt get execute for you too..

 

[renormalize]

@tellmesomething I have lightly edited your LaTeX to make it render (you should use fewer carriage returns to avoid excess whitespace):
$$R√2 tan \theta = x$$$$\frac {dx} {d\theta} = R√2 sec²\theta$$$$r= \frac { R√2} {cos \theta}$$$$dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 }$$$$\int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2}$$$$E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$$$E_{hori}= \frac {\lambda cos \phi } { 4π\epsilon R √2}$$$$E_{vert} =\frac { -\lambda } {4 π \epsilon R√6} (√3 + 1)$$$$E_{hori} =\frac {\lambda} { 4 π \epsilon R√3}$$

 

[tellmesomething]

@tellmesomething I have lightly edited your LaTeX to make it render (you should use fewer carriage returns to avoid excess whitespace):
$$R√2 tan \theta = x$$$$\frac {dx} {d\theta} = R√2 sec²\theta$$$$r= \frac { R√2} {cos \theta}$$$$dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 }$$$$\int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2}$$$$E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$$$E_{hori}= \frac {\lambda cos \phi } { 4π\epsilon R √2}$$$$E_{vert} =\frac { -\lambda } {4 π \epsilon R√6} (√3 + 1)$$$$E_{hori} =\frac {\lambda} { 4 π \epsilon R√3}$$

Thankyou!

 

[kuruman]

You have an expression for $dE$ in post #32 that looks like this

$$ dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 } $$

which is not the same as this expression for $dE$ in post #34.

The first expression has $\sec^2\theta$ in the numerator whilst the second has $\sec\theta.$ Which is correct?

When I provided you with a diagram in post #31 I did it so that you cab use it as a guide for redoing the integrals. Instead, it seems that you pasted your old solution. How can you find a mistake if you do that?

You show the integral $$\int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2}$$ Why are the limits of integration correct? If you integrate over $x$ the limits should be from $x=-R$ to $x=\infty.$ What would be the corresponding limits for $\theta$ as defined in my drawing?

For $E_{hori}$ you have angle $\phi$. What angle is that? There is no such angle in the drawing that I gave you.

Where is the setup for the $E_{vert}$ integral? How can I check your work if you don't show it and just write down the result?

 

[tellmesomething]

You have an expression for $dE$ in post #32 that looks like this

which is not the same as this expression for $dE$ in post #34.

View attachment 345313
The first expression has $\sec^2\theta$ in the numerator whilst the second has $\sec\theta.$ Which is correct?

When I provided you with a diagram in post #31 I did it so that you cab use it as a guide for redoing the integrals. Instead, it seems that you pasted your old solution. How can you find a mistake if you do that?

You show the integral $$\int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2}$$ Why are the limits of integration correct? If you integrate over $x$ the limits should be from $x=-R$ to $x=\infty.$ What would be the corresponding limits for $\theta$ as defined in my drawing?

For $E_{hori}$ you have angle $\phi$. What angle is that? There is no such angle in the drawing that I gave you.

Where is the setup for the $E_{vert}$ integral? How can I check your work if you don't show it and just write down the result?

The first one is correct I dont know why the latex reader missed the exponent...

I didnt understand what exactly you wanted me to redo since your diagram was some what similar to mine... I get it now

Phi is theta knot ... Since I didnt know how to type theta knot... Should have mentioned Sorry

Also im integrating over theta..im just realising my lower limit Should have been π/2 since its anti clockwise and negative..π/2 because if you go to infinity the angle between the normal from the point to the x axis and the line from the point to the end of inifinty would tend towards 90°


$$ E_{hori}= \int_{-π/2}^{\phi} \frac { \lambda sin \theta d\theta} { 4 π \epsilon R√2} $$

 

[tellmesomething]

You have an expression for $dE$ in post #32 that looks like this

which is not the same as this expression for $dE$ in post #34.

View attachment 345313
The first expression has $\sec^2\theta$ in the numerator whilst the second has $\sec\theta.$ Which is correct?

When I provided you with a diagram in post #31 I did it so that you cab use it as a guide for redoing the integrals. Instead, it seems that you pasted your old solution. How can you find a mistake if you do that?

You show the integral $$\int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2}$$ Why are the limits of integration correct? If you integrate over $x$ the limits should be from $x=-R$ to $x=\infty.$ What would be the corresponding limits for $\theta$ as defined in my drawing?

For $E_{hori}$ you have angle $\phi$. What angle is that? There is no such angle in the drawing that I gave you.

Where is the setup for the $E_{vert}$ integral? How can I check your work if you don't show it and just write down the result?

It looks like I just messed up the limits..sorry for the waste of time I think im all good the correct results should be

$$ E_{vert} = \frac { \lambda } { 4π \epsilon R√6} ( 1 + √3) $$

$$ E_{hori} = -\frac { \lambda} { 4 π \epsilon R√3} $$

 

[tellmesomething]

I looked at your drawing in post #10. The plane of the screen is the same as the plane of the triangle formed by the wire (along the x-axis), side $R\sqrt{2}$ and side $R\sqrt{3}$. The "horizontal" direction, unit vector $\mathbf {\hat h}$, is along the x-axis and the "vertical" direction, unit vector $\mathbf {\hat v}$, is along side $R\sqrt{2}$. So the contribution from the wire along the x-axis is $$\mathbf{E}_1=E_{hori}~\mathbf {\hat h}+E_{vert}~\mathbf {\hat v}.$$ View attachment 345295Now look at the drawing on the right. P is the point of interest (3,3,3). The plane of the red triangle is parallel to the $yz-$plane at distance $R$. Clearly $\mathbf {\hat h}=\mathbf {\hat x}$. Can you write $\mathbf {\hat v}$ as a linear combination of unit vectors $\mathbf {\hat y}$ and $\mathbf {\hat z}$? If yes, then a cyclic permutation of the axes should give you the contributions from the other two wires. There is also a symmetry argument that allows you to find the direction of the resultant field without adding any vectors. Can you formulate it?

I get the symmetry argument, we Can rename these axes and wed get the same magnitudes only different directions

$$ E_{x} = \frac { \lambda} {4π \epsilon R√3} (- \hat I ) + \frac { \lambda} { 8π \epsilon R} (√3 + 1) (\hat j ) + \frac { \lambda} { 8π \epsilon R} (√3 + 1) (\hat k) $$
$$ E_{y} = \frac { \lambda} {4π \epsilon R√3} (- \hat j ) + \frac { \lambda} { 8π \epsilon R} (√3 + 1) (\hat i) + \frac { \lambda} { 8π \epsilon R} (√3 + 1) (\hat k) $$
$$E_{z} = \frac { \lambda} {4π \epsilon R√3} (- \hat k) + \frac { \lambda} { 8π \epsilon R} (√3 + 1) (\hat j ) + \frac { \lambda} { 8π \epsilon R} (√3 + 1) (\hat i) $$

Net $$ \vec E = \frac { \lambda } { 4 π \epsilon R√3} ( 2+√3) (\hat i) +\frac { \lambda } { 4 π \epsilon R√3} ( 2+√3) (\hat j)+\frac { \lambda } { 4 π \epsilon R√3} ( 2+√3) (\hat k)$$

 

[kuruman]

Also im integrating over theta..im just realising my lower limit Should have been π/2 since its anti clockwise and negative..π/2 because if you go to infinity the angle between the normal from the point to the x axis and the line from the point to the end of inifinty would tend towards 90°

Not how it works. Look at my drawing. First of all the magnitude of $d\mathbf E$ is $$dE=\frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{(2R^2+x^2)}.$$Then $$dE_{vert}=dE\cos\theta=\frac{1}{4\pi\epsilon_0}\frac{\lambda R\sqrt{2}~dx}{(2R^2+x^2)^{3/2}}.$$ The vertical component is given by the integral $$E_{vert}=\frac{\lambda R\sqrt{2}}{4\pi\epsilon_0}\int_{-R}^{\infty}\frac{dx}{(2R^2+x^2)^{3/2}}.$$ At this point you can do the trigonometric substitution
Let $x=R\sqrt{2}\tan\theta$ in which case $~dx=\dfrac{R\sqrt{2}}{\cos^2\theta}d\theta~$ and $~(2R^2+x^2)^{3/2}=\cos^3\theta.$
Note that
when $x=-R$, $\theta=-\theta_0~;~~$ when $x=\infty$, $\theta=+\dfrac{\pi}{2}.$
The same limits of integration should be used for the $E_{hori}$ integral.

 

[kuruman]

Net $$ \vec E = \frac { \lambda } { 4 π \epsilon R√3} ( 2+√3) (\hat i) +\frac { \lambda } { 4 π \epsilon R√3} ( 2+√3) (\hat j)+\frac { \lambda } { 4 π \epsilon R√3} ( 2+√3) (\hat k)$$

And what direction is that? If you were to draw an arrow representing the net electric field at (3,3,3), how would you draw it?

 

[tellmesomething]

View attachment 345321Not how it works. Look at my drawing. First of all the magnitude of $d\mathbf E$ is $$dE=\frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{(2R^2+x^2)}.$$Then $$dE_{vert}=dE\cos\theta=\frac{1}{4\pi\epsilon_0}\frac{\lambda R\sqrt{2}~dx}{(2R^2+x^2)^{3/2}}.$$ The vertical component is given by the integral $$E_{vert}=\frac{\lambda R\sqrt{2}}{4\pi\epsilon_0}\int_{-R}^{\infty}\frac{dx}{(2R^2+x^2)^{3/2}}.$$ At this point you can do the trigonometric substitution
Let $x=R\sqrt{2}\tan\theta$ in which case $~dx=\dfrac{R\sqrt{2}}{\cos^2\theta}d\theta~$ and $~(2R^2+x^2)^{3/2}=\cos^3\theta.$
Note that
when $x=-R$, $\theta=-\theta_0~;~~$ when $x=\infty$, $\theta=+\dfrac{\pi}{2}.$
The same limits of integration should be used for the $E_{hori}$ integral.

Oh I sort of just accepted the fact that clockwise means positive anti means negative etc...I never really questioned it..I get the reasoning now and how the above assumption could lead to wrong. Thankss..

 

[tellmesomething]

And what direction is that? If you were to draw an arrow representing the net electric field at (3,3,3), how would you draw it?

positive for all three..? Outward from a 2d page with a cuboid diagram... Like in the direction of the outward diagonal...

 

[tellmesomething]

positive for all three..? Outward from a 2d page with a cuboid diagram... Like in the direction of the outward diagonal...

 

[kuruman]

positive for all three..? Outward from a 2d page with a cuboid diagram... Like in the direction of the outward diagonal...

Exactly but cube of side $R$, not cuboid. The symmetry argument is that when you look down the cube diagonal, you see a charge distribution that looks the same if you rotate by 120° about this cube diagonal. If the net field had a component perpendicular to the diagonal (red arrow, figure on the right), it would rotate with it. This creates the paradox that identical charge distributions do not give rise to identical net fields. Therefore there is no component of the net field perpendicular to the cube diagonal.

Once you find the field due to the wire along the x-axis, $\mathbf E^{(x)}$ you can find its component along the diagonal $E_{diag}=\mathbf E^{(x)}\cdot \left(\dfrac{\hat i+\hat j+\hat k}{\sqrt{3}}\right)$ and multiply the result by 3.

 

[tellmesomething]

View attachment 345326Exactly but cube of side $R$, not cuboid. The symmetry argument is that when you look down the cube diagonal, you see a charge distribution that looks the same if you rotate by 120° about this cube diagonal. If the net field had a component perpendicular to the diagonal (red arrow, figure on the right), it would rotate with it. This creates the paradox that identical charge distributions do not give rise to identical net fields. Therefore there is no component of the net field perpendicular to the cube diagonal.

Once you find the field due to the wire along the x-axis, $\mathbf E^{(x)}$ you can find its component along the diagonal $E_{diag}=\mathbf E^{(x)}\cdot \left(\dfrac{\hat i+\hat j+\hat k}{\sqrt{3}}\right)$ and multiply the result by 3.

Ad you might've already guessed im very slow. I dont get you after "rotate by 120° about this cube diagonal". Your telling me that if I rotate the cube about its diagonal I would get a similar charge distribution yes...but shouldn't the net field still point towards the same direction...instead here it rotate E about 120°....and after that the perpendicular argument went over my head..

 

[kuruman]

Maybe I didn't explain myself clearly. You don't rotate the cube, you rotate all three lines of charge by 120° about an axis that is the line joining the origin of coordinates and point (3,3,3). When you perform this rotation, the charge distribution looks exactly the same after the rotation as before. Therefore, the net field must look the same as before. Now suppose that in the "before" picture, figure (A) post #46, the net field has a component perpendicular to the rotation axis as indicated by the red arrow. If I rotate by 120°, the red arrow should be rotated together with the distribution that created it and you would have figure (B). That cannot be because we have identical charge distributions producing non-identical net fields. The only way out is to have no perpendicular component of the net field.

 

[tellmesomething]

Maybe I didn't explain myself clearly. You don't rotate the cube, you rotate all three lines of charge by 120° about an axis that is the line joining the origin of coordinates and point (3,3,3). When you perform this rotation, the charge distribution looks exactly the same after the rotation as before. Therefore, the net field must look the same as before. Now suppose that in the "before" picture, figure (A) post #46, the net field has a component perpendicular to the rotation axis as indicated by the red arrow. If I rotate by 120°, the red arrow should be rotated together with the distribution that created it and you would have figure (B). That cannot be because we have identical charge distributions producing non-identical net fields. The only way out is to have no perpendicular component of the net field.

I see I think I get it a little ill have to keep rereading it to sort of absorb it fully....last question i promise, the arrow in your diagram in post #46 represents the perpendicular to the rotation axis..?

 

[kuruman]

I see I think I get it a little ill have to keep rereading it to sort of absorb it fully....last question i promise, the arrow in your diagram in post #46 represents the perpendicular to the rotation axis..?

It represents the net field with a perpendicular component that cannot exist. If it did, it would have to rotate to new position when the charge distribution is rotated. The idea is simple. Look at figure (A) on the right. The lines of charge are colored to tell them apart but they have identical charge per unit length length $\lambda.$ If I rotate (A) by 120° about an axis that is perpendicular to the screen and connects point (3,3,3) with the origin, I get figure (B). The charge distribution in (B) is identical to (A). This means that the net field everywhere in space in (B) must be identical to (A). The only way that the net field can look the same before and after the rotation is if its perpendicular component is zero because this component rotates with the distribution while the component along the axis does not.

 

[tellmesomething]

View attachment 345350It represents the net field with a perpendicular component that cannot exist. If it did, it would have to rotate to new position when the charge distribution is rotated. The idea is simple. Look at figure (A) on the right. The lines of charge are colored to tell them apart but they have identical charge per unit length length $\lambda.$ If I rotate (A) by 120° about an axis that is perpendicular to the screen and connects point (3,3,3) with the origin, I get figure (B). The charge distribution in (B) is identical to (A). This means that the net field everywhere in space in (B) must be identical to (A). The only way that the net field can look the same before and after the rotation is if its perpendicular component is zero because this component rotates with the distribution while the component along the axis does Not.

Okay I get it I think i'll put this to test now. Thankyou so much for your immense help...You've helped with over 30 posts and have been very patient and helpful... Thankyou