Find the electric field at a point in 3 dimensional space

[tellmesomething]

Homework Statement

3 semi infinite rods are placed along x, y and z axis such that the points on them obey x > 0,y>0 and z>0 and their common end is origin. Find Electric field at (R,R,R)

Relevant Equations

None

I tried resolving the semi infinite rods into arcs of 90 degree each placed on the three axes but that doesnt take me anywhere....
Alternatively I tried finding out the field at the point due to each rod but im unable to find the perpendicular distance from the point to the rod...I dont think such a distance exists..I know I should use integration but im having a hard time doing anything in 3d. Can someone give me a starter ..

 

[kuruman]

I dont think such a distance exists.

Sure it exists. Consider the rods as three mutually perpendicular Cartesian axes. The point (R,R,R) is described by vector $\mathbf R=R~\mathbf {\hat x}+R~\mathbf {\hat y}+R~\mathbf {\hat z}.$ What is the perpendicular distance to, say, the rod along the x-axis?

 

[tellmesomething]

The

Sure it exists. Consider the rods as three mutually perpendicular Cartesian axes. The point (R,R,R) is described by vector $\mathbf R=R~\mathbf {\hat x}+R~\mathbf {\hat y}+R~\mathbf {\hat z}.$ What is the perpendicular distance to, say, the rod along the x-axis?

For it to be perpendicular to the x axis from the point, it would have to be parallel to either the y or the z axis no?

 

[kuruman]

For it to be perpendicular to the x axis from the point, it would have to be parallel to either the y or the z axis no?

Why? If you have a single x-axis in 3D space can you not always draw a perpendicular to it from any point in space? You can do that first and then add the y and z axes in a plane perpendicular to the x-axis any way you please.

I suggest that you first make a good drawing of how one gets to (R, R, R) from the origin in Cartesian coordinates. Then do some trig.

 

[tellmesomething]

Why? If you have a single x-axis in 3D space can you not always draw a perpendicular to it from any point in space? You can do that first and then add the y and z axes in a plane perpendicular to the x-axis any way you please.

I suggest that you first make a good drawing of how one gets to (R, R, R) from the origin in Cartesian coordinates. Then do some trig.

Okay so it took a while but I did do this I found out the perpendicular distance from x axis to be R√2. Now im having trouble integrating since I am unaware what to put in for the lower limit, the upper limit is 90 but the lower limit ends up being an unknown..s

 

[kuruman]

Okay so it took a while but I did do this I found out the perpendicular distance from x axis to be R√2. Now im having trouble integrating since I am unaware what to put in for the lower limit, the upper limit is 90 but the lower limit ends up being an unknown..s

The $R\sqrt{2}$ answer is correct. If you want additional help with your integral, please provide the drawing on which you based it plus the expression of the integral itself. I cannot read your mind and I won't even try.

 

[tellmesomething]

The $R\sqrt{2}$ answer is correct. If you want additional help with your integral, please provide the drawing on which you based it plus the expression of the integral itself. I cannot read your mind and I won't even try.

 

[tellmesomething]

View attachment 345243

I think ill type it out too its not very clear please wait for some time

 

[tellmesomething]

The $R\sqrt{2}$ answer is correct. If you want additional help with your integral, please provide the drawing on which you based it plus the expression of the integral itself. I cannot read your mind and I won't even try.

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

 

[tellmesomething]

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

The angle between the perpendicular line and the line joining the point to the origin is $\phi$

 

[haruspex]

The angle between the perpendicular line and the line joining the point to the origin is

So $\cos(\phi)$ and $\sin(\phi)$ are… ?

 

[kuruman]

Also keep in mind that the component you call $E_{hori}$ is along the x-axis, however component $E_{vert}$, although perpendicular to the $x$- axis, is not parallel to either $y$ or $z$. You need to be careful about how you add the contributions from the other two wires. Alternatively, you can use a symmetry argument.

 

[tellmesomething]

So $\cos(\phi)$ and $\sin(\phi)$ are… ?

Right the former is $\frac {√2} { √3}$ and the latter becomes $\frac {1} {√3}$

 

[tellmesomething]

Also keep in mind that the component you call $E_{hori}$ is along the x-axis, however component $E_{vert}$, although perpendicular to the $x$- axis, is not parallel to either $y$ or $z$. You need to be careful about how you add the contributions from the other two wires. Alternatively, you can use a symmetry argument.

How did you figure out the horizontal component is parallel to the x axis ?

 

[tellmesomething]

Also keep in mind that the component you call $E_{hori}$ is along the x-axis, however component $E_{vert}$, although perpendicular to the $x$- axis, is not parallel to either $y$ or $z$. You need to be careful about how you add the contributions from the other two wires. Alternatively, you can use a symmetry argument.

I was thinking of finding the magnitude of net E from the two components and doing the same for the other axes and since the net E from the three axes must be mutually perpendicular(?) I am gonna use that to find the total E....doe this sound wrong??

 

[haruspex]

the net E from the three axes must be mutually perpendicular

For each axis, there is an E component normal to that axis and one parallel to it.
The parallel three will be normal to each other, but what about the first three? Is the line x=y, z=0 normal to x=z, y=0?
Even if both triplets were to consist of three orthogonal vectors, would that prove the three resultants are mutually orthogonal?

 

[tellmesomething]

For each axis, there is an E component normal to that axis and one parallel to it.
The parallel three will be normal to each other, but what about the first three? Is the line x=y, z=0 normal to x=z, y=0?
Even if both triplets were to consist of three orthogonal vectors, would that prove the three resultants are mutually orthogonal?

Oh. That makes sense. Though I dont understand how you concluded that one component will be parallel to the axis, is it something you're immediately able to see? Or did you like make a diagram cause its not obvious to me yet

 

[haruspex]

Oh. That makes sense. Though I dont understand how you concluded that one component will be parallel to the axis, is it something you're immediately able to see? Or did you like make a diagram cause its not obvious to me yet

The field at a point due to a given wire will lie in a plane containing that wire. It can therefore be decomposed into a component parallel to the wire and a component, in that plane, normal to the wire.
It's not that one component will necessarily be parallel to the wire, it’s that it can be decomposed so that one is.

 

[tellmesomething]

It's not that one component will necessarily be parallel to the wire, it’s that it can be decomposed so that one is.

I didnt get this can you elaborate? To me both of these sound like the same thing

 

[tellmesomething]

For each axis, there is an E component normal to that axis and one parallel to it.
The parallel three will be normal to each other, but what about the first three? Is the line x=y, z=0 normal to x=z, y=0?
Even if both triplets were to consist of three orthogonal vectors, would that prove the three resultants are mutually orthogonal?

Also what method do you suggest then to find the total field due to all the three rods at point (R,R,R)..?

 

[haruspex]

Also what method do you suggest then to find the total field due to all the three rods at point (R,R,R)..?

In post #10 you found two components that add to make the field due to one wire. Suppose that is the wire along the x axis. What are those two components written in terms of $\hat x, \hat y, \hat z$?

 

[tellmesomething]

In post #10 you found two components that add to make the field due to one wire. Suppose that is the wire along the x axis. What are those two components written in terms of $\hat x, \hat y, \hat z$?

OK. I can try but i'll definitely take a lot of time..

 

[haruspex]

OK. I can try but i'll definitely take a lot of time..

It's quite straightforward. One of them contributes only to $E_x$, while the other contributes to $E_y$ and $E_z$.

 

[tellmesomething]

It's quite straightforward. One of them contributes only to $E_x$, while the other contributes to $E_y$ and $E_z$.

Sorry for the late reply I keep trying then giving up. I have some questions for a normal 3D vector I figured out:

The magnitude of the components are the cosine of the angle the vector makes with the axis times the magnitude of the resultant. Thats ok

But for a vector like this which starts from the middle of the axis and goes till a point in space

How do I resolve this? I understand this is a sum of two vectors so do I find those two vectors and their components...?

I understand that you're trying to stress on the fact that its quite easy to figure it out but im having a hard time convincing myself so imt trying to prove everything... Sorry

 

[kuruman]

How did you figure out the horizontal component is parallel to the x axis ?

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 $$\mathbf{E}_1=E_{hori}~\mathbf {\hat h}+E_{vert}~\mathbf {\hat v}.$$

Now 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?

 

[kuruman]

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

The negative sign bothers me. For positive $\lambda$ the vertical component $E_{vert}~\mathbf{\hat v}$ must point away from the wire in the direction of $~\mathbf{\hat v}$. The negative sign implies "towards the wire."

 

[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?

It all makes sense. All I needed to do was use technology to make a diagram the diagrams on my 2 dimensional notebook page made it more confusing. If you can please let me know what you used to create this diagram... I thought of paint 3d..I have the angle from the diagram and the components but im sure I am not wise enough to find the symmetry argument so ill not look that way. About post 26 im sure its a mathematical error I'll fix that asap. Thanks a ton

 

[tellmesomething]

It all makes sense. All I needed to do was use technology to make a diagram the diagrams on my 2 dimensional notebook page made it more confusing. If you can please let me know what you used to create this diagram... I thought of paint 3d..I have the angle from the diagram and the components but im sure I am not wise enough to find the symmetry argument so ill not look that way. About post 26 im sure its a mathematical error I'll fix that asap. Thanks a ton

Not a wise comment sorry its pretty evident its just a 2d diagram..I just need to be clear in my work

 

[tellmesomething]

The negative sign bothers me. For positive $\lambda$ the vertical component $E_{vert}~\mathbf{\hat v}$ must point away from the wire in the direction of $~\mathbf{\hat v}$. The negative sign implies "towards the wire."

Okay so I did this again... Im getting the same expression which is definitely wrong..I know this is not a check my work site and Youve already helped me directly by literally drawing me a diagram but can you check the integration and give me a hint as to where i went wrong..?

 

[tellmesomething]

Okay so I did this again... Im getting the same expression which is definitely wrong..I know this is not a check my work site and Youve already helped me directly by literally drawing me a diagram but can you check the integration and give me a hint as to where i went wrong..?

Im my opinion the E sin theta component is also wrong since its positive but if you analyse the sum of the components along the x axis should point in the negative x direction..

 

[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.