Electric Potential Field Calculation

In summary: You have to calculate the potential field at the point (0,0,h) using the equation given in the homework statement.
  • #1
nownow
4
2
Homework Statement
A finite line of charges has been bent into the shape shown in the figure below (2 orthogonal segments and an arc of a circle). Knowing the charge density in the line is given by ρl (C/m) get the electric potential field at the point (0,0,h). h>a
Relevant Equations
electric potential field equation
1655672351756.png

I've already tried to calculate the potential with respect to the 3 segments and then apply superposition (V1+V2+V3). However, I was not very successful. My error I think is in the calculation of the radii, mainly of the line segment that is on the z axis. Can anybody help me? I need some light on this question, please
 
Physics news on Phys.org
  • #2
nownow said:
Homework Statement:: A finite line of charges has been bent into the shape shown in the figure below (2 orthogonal segments and an arc of a circle). Knowing the charge density in the line is given by ρl (C/m) get the electric potential field at the point (0,0,h). h>a
Relevant Equations:: electric potential field equation

View attachment 303063
I've already tried to calculate the potential with respect to the 3 segments and then apply superposition (V1+V2+V3). However, I was not very successful. My error I think is in the calculation of the radii, mainly of the line segment that is on the z axis. Can anybody help me?
We can help you if you show us what you did. Show us your work and where you think you went wrong. Specifically, what radii are you talking about? There is only one radius and that is associated with the quarter-circle. Your superposition approach is correct.
 
  • Like
Likes bob012345
  • #3
kuruman said:
We can help you if you show us what you did. Show us your work and where you think you went wrong. Specifically, what radii are you talking about? There is only one radius and that is associated with the quarter-circle.
It's not multiple rays. I expressed myself badly, sorry. However, on to my calculations:

$$V_1 = \int_{0}^{a}\frac{\rho_l}{4\pi \varepsilon_0\sqrt{x^2+h^2} }dx$$
$$V_2 = \int_{0}^{a}\frac{\rho_l}{4\pi \varepsilon_0(h-z) }dz$$
$$V_3 = \int_{0}^{\frac{\pi }{2}}\frac{\rho_la}{4\pi \varepsilon_0\sqrt{a^2+h^2} }d\phi$$

Is what I did correct?
 
  • #4
what is the angle between the x and y axis?
 
  • #5
Orodruin said:
what is the angle between the x and y axis?
The angle is ##\frac{\pi }{2}##
 
  • #6
nownow said:
The angle is ##\frac{\pi }{2}##
So why does your integral specify integration between zero and ##\pi/4##?

Also, editing out an error from your post after getting a reply is not good form.
 
  • #7
I ended up realizing my error shortly after sending the answer. Also, I didn't edit my answer because of your question. Anyway, I apologize for my mistake and I didn't know about this no edit practice, I'm new here. If you want I can edit the question to what it was before, no problem. I just want to clear my doubt on the question. Again sorry for the edit. Thanks
 
  • Like
Likes PeroK and berkeman
  • #8
That looks about right. What do you get when you redo the integrals as you have them in post #4 #3? Is your answer the same as you got before?
 
Last edited:
  • Like
Likes Delta2
  • #9
post #3 setups correctly the integrals. Now the OP has to calculate them correctly.
 
  • Like
Likes kuruman
  • #10
BTW there was another thread, I think from the same OP that the source charge density was portion of a cylindrical surface, so we would have to do a surface integral, I had post to that thread but can't find it anymore, what happened?
 

Related to Electric Potential Field Calculation

1. What is an electric potential field?

An electric potential field is a region in space where a charged particle experiences a force due to the presence of other charged particles. It is created by the distribution of electric charges in space and is represented by a vector field.

2. How is an electric potential field calculated?

An electric potential field is calculated by using the formula V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the particle, and r is the distance from the particle. This formula takes into account the strength of the electric charge and the distance from it.

3. What is the unit of measurement for electric potential field?

The unit of measurement for electric potential field is volts (V). This is the same unit used for electric potential, as they are directly related. One volt is equivalent to one joule per coulomb.

4. How does the electric potential field affect the motion of charged particles?

The electric potential field affects the motion of charged particles by exerting a force on them. The direction of the force is determined by the direction of the electric field lines, which point from positive to negative charges. This force can either attract or repel charged particles, causing them to accelerate or decelerate accordingly.

5. What factors can affect the strength of an electric potential field?

The strength of an electric potential field can be affected by the amount and distribution of electric charges in the space. The distance between the charges also plays a role, as the electric field strength decreases with distance. Additionally, the medium through which the charges are moving can also impact the strength of the electric potential field.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
Replies
1
Views
223
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
859
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
278
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
23
Views
423
  • Introductory Physics Homework Help
Replies
3
Views
846
Back
Top