Electric field due to semi-circular wire at a distance

  • #1
Gijs

Homework Statement


A semi-circular wire containing a total charge Q which is uniformly distribute over the wire in the x-y plane. the semi-circle has a radius a and the origin is the center of the circle.

Now I want to calculate the electric field at a point located on at distance h on the z-axis. What is the magnitude of this electric field, and is this field only in the z direction?

Homework Equations



Coulomb's Law

The Attempt at a Solution



dE=dQ/(4πε√h^2+a^2)
dEz=dE cosθ = dQ/(4πε√(h^2+a^2))(h/(√(h^2+a^2))
E=Qh/(4πε(h^2+a^2)^(3/2))
[/B]
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,734
3,097
Hello and welcome to PF!

Will the total field on the z axis have only a z component?
 
  • #3
Gijs
Thanks! I guess not since it not all θ components will cancel out. This is what I’m confued about. Can you explain how I could solve this?
 
  • #4
TSny
Homework Helper
Gold Member
12,734
3,097
Thanks! I guess not since it not all θ components will cancel out.
Right. The total field will have a component, E, that is perpendicular to the z axis .
This is what I’m confused about. Can you explain how I could solve this?
From symmetry, you should be able to see the direction of E. The problem did not state if the orientation of the x and y axes are given. If not, orient the axes to take advantage of the symmetry.

For an arbitrary element of charge dQ, you will need to figure out how to use trig to project out the relevant component of the field due to dQ. A good diagram will be helpful.
 
  • #5
Gijs
Right. The total field will have a component, E, that is perpendicular to the z axis .

From symmetry, you should be able to see the direction of E. The problem did not state if the orientation of the x and y axes are given. If not, orient the axes to take advantage of the symmetry.

For an arbitrary element of charge dQ, you will need to figure out how to use trig to project out the relevant component of the field due to dQ. A good diagram will be helpful.
So if z=0 then I get 2kQ/(a^2) but when the hight is h I get 2kQh/sqrt(a^2 + h^2) is this correct?
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,317
1,007
So if z=0 then I get 2kQ/(a^2) but when the hight is h I get 2kQh/sqrt(a^2 + h^2) is this correct?
No.

For one thing, that answer has the wrong units.

What is the magnitude of the electric field at that point?
 
  • #7
Gijs
No.

For one thing, that answer has the wrong units.

What is the magnitude of the electric field at that point?
I’m not sure what you mean? This is the magnitude I have calculated. Can you give me some directions how to get to the right anwser? And what should be the direction of the electric field in in point h in catesian coordinates?
 
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,317
1,007
I’m not sure what you mean? This is the magnitude I have calculated. Can you give me some directions how to get to the right anwser? And what should be the direction of the electric field in in point h in catesian coordinates?
Your previous post has a number of errors.
So if z=0 then I get 2kQ/(a^2) but when the hight is h I get 2kQh/sqrt(a^2 + h^2) is this correct?
At z = 0, the z component of electric field is zero. The horizontal component is nearly what you have but it's ##\displaystyle \ \frac {2kQ} {\pi a^2} \ ##.

You also have some problems in your attempt at a solution in post #1.

The Attempt at a Solution



dE=dQ/(4πε√h^2+a^2)[/B]
That's not quite right.
##\displaystyle dE=\frac{dQ}{4\pi\varepsilon_0 R^2}=\frac{dQ}{4\pi\varepsilon_0 (h^2+a^2)}##
dEz=dE cosθ = dQ/(4πε√(h^2+a^2))(h/(√(h^2+a^2))
Except for the part I highlighted in red, that may be correct, depending upon how you have defined θ.

As it turns out, I think the following expression is correct for the z component of electric field at z = h on the z axis .
E=Qh/(4πε(h^2+a^2)^(3/2))
 
Last edited:
  • #9
Gijs
Your previous post has a number of errors.

At z = 0, the z component of electric field is zero. The horizontal component is nearly what you have but it's ##\displaystyle \ \frac {2kQ} {\pi a^2} \ ##.

You also have some problems in your attempt at a solution in post #1.
That's not quite right.
##\displaystyle dE=\frac{dQ}{4\pi\varepsilon_0 R^2}=\frac{dQ}{4\pi\varepsilon_0 (h^2+a^2})##
Except for the part I highlighted in red, that may be correct, depending upon how you have defined θ.

As it turns out, I think the following expression is correct for the z component of electric field at z = h on the z axis .
Thanks a lot for you help! However I am still confused about the direction of the electric field. So I added a drawing to clarify. I calculated the electric field in the z component but from the image is clear that this is not the only component. What geometry did I mis and how should I calculated te E field in point P? Thanks again!
 

Attachments

  • #10
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,317
1,007
Thanks a lot for you help! However I am still confused about the direction of the electric field. So I added a drawing to clarify. I calculated the electric field in the z component but from the image is clear that this is not the only component. What geometry did I mis and how should I calculated te E field in point P? Thanks again!
It certainly does help to have a good clear image of a sketch of the situation. Here is the one you linked.
c1c2dab3-32cf-4f0f-9022-90e4f4eec6e0-jpeg.jpg

The expression you have for dE in your third line is in error. It should not have the square root. The third line should be:
##\displaystyle =\frac{dQ}{4\pi\varepsilon_0 (h^2+a^2)}\cdot\frac{h}{\sqrt{h^2+a^2}}##​

A similar problem is solved in the following thread. The choice of coordinates is different than what you have.
Electric field of a semi circle ring

The choice of coordinates is different than what you have.
 

Attachments

  • #11
Gijs
You are right! I have seen that topic however that is containing a full circle where the there is only a component in the z direction.
What I have so far is the following:

$$
\begin{align}
\vec{r}_{p}&=\vec{ha_{z}}\\
\vec{r}_{s}&=a\vec{a_{x}} cos\phi +a\vec{a_{y}} sin\phi\\
dE&=\frac{dQ(\vec{r}_{p}-\vec{r}_{s})}{4\pi \epsilon_{0}\left |\vec{r}_{p}-\vec{r}_{s} \right |^{3}}\\
dE&=\frac{dQ \, \vec{ha_{z}}-(a\vec{a_{x}}cos\phi +a\vec{a_{y}} sin\phi) } {4\pi \epsilon_{0}(a^{2}+h^{2} )^{3/2}}\\
dQ&=\rho _{L}ad\phi \\
E&=\int_{0}^{\pi}\frac{dQ \, \vec{ha_{z}}-(a\vec{a_{x}}cos\phi +a\vec{a_{y}} sin\phi) } {4\pi \epsilon_{0}(a^{2}+h^{2} )^{3/2}}d\phi\\
\end{align}$$

However I am not sure how I should integrate this in order to find the contribution in every direction?
 
  • #12
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,317
1,007
You are right! I have seen that topic however that is containing a full circle where the there is only a component in the z direction.
What I have so far is the following:

$$
\begin{align}
\vec{r}_{p}&=\vec{ha_{z}}\\
\vec{r}_{s}&=a\vec{a_{x}} cos\phi +a\vec{a_{y}} sin\phi\\
dE&=\frac{dQ(\vec{r}_{p}-\vec{r}_{s})}{4\pi \epsilon_{0}\left |\vec{r}_{p}-\vec{r}_{s} \right |^{3}}\\
dE&=\frac{dQ \, \vec{ha_{z}}-(a\vec{a_{x}}cos\phi +a\vec{a_{y}} sin\phi) } {4\pi \epsilon_{0}(a^{2}+h^{2} )^{3/2}}\\
dQ&=\rho _{L}ad\phi \\
E&=\int_{0}^{\pi}\frac{dQ \, \vec{ha_{z}}-(a\vec{a_{x}}cos\phi +a\vec{a_{y}} sin\phi) } {4\pi \epsilon_{0}(a^{2}+h^{2} )^{3/2}}d\phi\\
\end{align}$$

However I am not sure how I should integrate this in order to find the contribution in every direction?
(Actually that link IS for a semi-circle of uniform charge. )

I don't entirely understand your notation.

From what axis is angle ##\ \phi\ ## measured ?

What is meant by ##\ \vec{a_{x}},\,\vec{a_{y}},\,\vec{a_{z}}\ ## ?
 
Last edited:
  • #13
Gijs
(Actually that link IS for a semi-circle of uniform charge. )

I don't entirely understand your notation.

From what axis is angle ##\ \phi\ ## measured ?

What is meant by ##\ \vec{a_{x}},\,\vec{a_{y}},\,\vec{a_{z}}\ ## ?
Thanks for pointing me to the other topic again. That solved my question and made it clear. Again thanks for your help!
 

Related Threads on Electric field due to semi-circular wire at a distance

Replies
5
Views
6K
Replies
0
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
18
Views
2K
  • Last Post
Replies
6
Views
638
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
21
Views
9K
Replies
3
Views
1K
Top