Find the electric field at a point in 3 dimensional space

  • #1
tellmesomething
198
20
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
%2Fstorage%2Femulated%2F0%2FDCIM%2FScreenshots%2FIMG_20240514_010636.jpg


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 ..
 
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  • #2
tellmesomething said:
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?
 
  • #3
kuruman said:
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?
 
  • #4
tellmesomething said:
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.
 
  • #5
kuruman said:
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
 
  • #6
tellmesomething said:
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.
 
  • #7
kuruman said:
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.
Screenshot_2024-05-14-23-00-55-139_com.google.android.gm.jpg
 
  • #8
tellmesomething said:
I think ill type it out too its not very clear please wait for some time
 
  • #9
kuruman said:
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.
$$ R√2 tan \theta = x $$
$$ \frac {dx} {d\theta} = R√2 sec²\theta $$
$$ dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 } $$
$$ \int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2} $$
$$ E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$
$$ E_{hori}= \frac {\lambda cos \phi } { 4π\epsilon R √2} $$
 
Last edited:
  • #10
tellmesomething said:
$$ R√2 tan \theta = x $$
$$ \frac {dx} {d\theta} = R√2 sec²\theta $$
$$ dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 } $$
$$ \int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2} $$
$$ E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$
$$ E_{hori}= \frac {\lambda cos \phi } { 4π\epsilon R √2} $$
Screenshot_2024-05-14-23-32-41-836_com.miui.gallery.jpg

The angle between the perpendicular line and the line joining the point to the origin is ## \phi ##
 
  • #11
tellmesomething said:
The angle between the perpendicular line and the line joining the point to the origin is
So ##\cos(\phi)## and ##\sin(\phi)## are… ?
 
  • #12
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.
 
  • #13
haruspex said:
So ##\cos(\phi)## and ##\sin(\phi)## are… ?
Right the former is ## \frac {√2} { √3} ## and the latter becomes ## \frac {1} {√3} ##
 
  • #14
kuruman said:
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 ?
 
  • #15
kuruman said:
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??
 
  • #16
tellmesomething said:
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?
 
  • #17
haruspex said:
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
 
  • #18
tellmesomething said:
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.
 
  • #19
haruspex said:
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
 
  • #20
haruspex said:
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)..?
 
  • #21
tellmesomething said:
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##?
 
  • #22
haruspex said:
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..
 
  • #23
tellmesomething said:
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##.
 
  • #24
haruspex said:
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:
Screenshot_2024-05-15-17-35-36-183_com.miui.gallery.jpg

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
Screenshot_2024-05-15-17-36-16-644_com.miui.gallery.jpg

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
 
  • #25
tellmesomething said:
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 $$\mathbf{E}_1=E_{hori}~\mathbf {\hat h}+E_{vert}~\mathbf {\hat v}.$$
Tri-wire.png
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?
 
Last edited:
  • #26
tellmesomething said:
$$ E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$
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."
 
  • #27
kuruman said:
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
 
  • #28
tellmesomething said:
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
 
  • #29
kuruman said:
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..?
 
  • #30
tellmesomething said:
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..
 
  • #31
tellmesomething said:
... 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.


Tri-wire_2.png
 
  • #32
kuruman said:
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
$$ R√2 tan \theta = x $$

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

$$ r= \frac { R√2} {cos \theta} $$

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

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

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

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

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

$$ E_{hori} =\frac {\lambda} { 4 π \epsilon R√3} $$
 
  • #33
tellmesomething said:
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
 
  • #34
tellmesomething said:
Following this
$$ R√2 tan \theta = x $$

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

$$ r= \frac { R√2} {cos \theta} $$

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

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

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

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

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

$$ E_{hori} =\frac {\lambda} { 4 π \epsilon R√3} $$
IMG_20240516_003839.jpg

This is what it should be showing incase it didnt get execute for you too..
 
  • #35
@tellmesomething I have lightly edited your LaTeX to make it render (you should use fewer carriage returns to avoid excess whitespace):
$$R√2 tan \theta = x$$$$\frac {dx} {d\theta} = R√2 sec²\theta$$$$r= \frac { R√2} {cos \theta}$$$$dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 }$$$$\int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos \theta } { 4π \epsilon R√2}$$$$E_{vert} = \frac { - \lambda } { 4 π \epsilon R√2} ( 1 + sin \phi) $$$$E_{hori}= \frac {\lambda cos \phi } { 4π\epsilon R √2}$$$$E_{vert} =\frac { -\lambda } {4 π \epsilon R√6} (√3 + 1)$$$$E_{hori} =\frac {\lambda} { 4 π \epsilon R√3}$$
 
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