Semi-circle linear charge Electric Field

[RJLiberator]

Homework Statement

A semi-circle of radius R has a charge Q uniformly distributed over its length, which provides a line charge density λ. Determine E at the origin.

Homework Equations

The Attempt at a Solution

https://www.physicsforums.com/attachments/105239

I can tell by argument of symmetry that the Electric field will be pointing in +y direction.

If we take a sliver of the charge, call it dq, we will calculate the Electric field.

[tex] E = \left(\frac{1}{4πε_0}\right)\left(\frac{dq}{R^2}\right)sin(θ) [/tex]

Also: [tex] \left(\frac{dq}{dθ}\right) = \left(\frac{Q}{π}\right) [/tex]

so: [tex] dq = \left(\frac{Qdθ}{π}\right) [/tex]

Now:

[tex] E = \left(\frac{1}{4πε_0}\right)\left(\frac{Qsin(θ)}{πR^2}\right)dθ [/tex]

After calculating the integral from θ = 0 to θ = π I get the following answer:[tex] E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^2}\right)\hat{y} [/tex]

Does this work appear to be correct? I want to make sure I get this easy material down before I go to the more difficult material in electromagnetism. If so, is it fair to say that the linear charge density typically symbolized as λ is equal to Q/π in this problem. Q is the charge in coulombs and pi is the length of the semicircle. So I could represent my answer as:

[tex] E = \left(\frac{1}{2πε_0}\right)\left(\frac{λ}{R^2}\right)\hat{y} [/tex]

 

[blue_leaf77]

https://www.physicsforums.com/attachments/105239/

I cannot open the file.

Does this work appear to be correct?

Regardless of the missing picture, your final answer is correct.

is it fair to say that the linear charge density typically symbolized as λ is equal to Q/π in this problem.

No, that cannot be a linear charge density since its dimension is the same as that of the charge. Note that the circumference of a semicircle must contain its radius.

 

[SammyS]

Homework Statement

A semi-circle of radius R has a charge Q uniformly distributed over its length, which provides a line charge density λ. Determine E at the origin.

Homework Equations

The Attempt at a Solution

https://www.physicsforums.com/attachments/105239

I can tell by argument of symmetry that the Electric field will be pointing in +y direction.

If we take a sliver of the charge, call it dq, we will calculate the Electric field.

[tex] E = \left(\frac{1}{4πε_0}\right)\left(\frac{dq}{R^2}\right)sin(θ) [/tex]

Also: [tex] \left(\frac{dq}{dθ}\right) = \left(\frac{Q}{π}\right) [/tex]

so: [tex] dq = \left(\frac{Qdθ}{π}\right) [/tex]

Now:

[tex] E = \left(\frac{1}{4πε_0}\right)\left(\frac{Qsin(θ)}{πR^2}\right)dθ [/tex]

After calculating the integral from θ = 0 to θ = π I get the following answer:[tex] E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^2}\right)\hat{y} [/tex]

Does this work appear to be correct? I want to make sure I get this easy material down before I go to the more difficult material in electromagnetism. If so, is it fair to say that the linear charge density typically symbolized as λ is equal to Q/π in this problem. Q is the charge in coulombs and pi is the length of the semicircle. So I could represent my answer as:

[tex] E = \left(\frac{1}{2πε_0}\right)\left(\frac{λ}{R^2}\right)\hat{y} [/tex]

The radius is not 1, is it?

The charge is spread over the semicircle. What is 1/2 the circumference of a circle of radius, R ?

Also, missing from the problem statement is the location and orientation of the semi-circle. (I could not access the image.)

 

[RJLiberator]

Hey guys,

Sorry about the image not showing -- it was an accidental double post.

I see your point on the radius. I watched a video on this problem and did not take into consideration that the radius of that video = 1, while my radius = R. So the Linear charge density would be Q/(pi*R), is this now correct?

Due to this fact, I will have to change my answer slightly to:

[tex]E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^3}\right)\hat{y}[/tex]

 

[blue_leaf77]

So the Linear charge density would be Q/(pi*R), is this now correct?

It's now correct.

Due to this fact, I will have to change my answer slightly to:

E=(12πε0)(QπR3)^yE=(12πε0)(QπR3)y^​

Why would there be an extra ##R##? Your previous answer (2nd equation from the last in post#1) is already correct.

 

[RJLiberator]

If linear charge density
[tex] λ = \left(\frac{Q}{πR}\right) [/tex]

Then the final answer :

[tex]E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^2}\right)\hat{y} [/tex]

which is:
[tex]E = \left(\frac{1}{2πε_0}\right)\left(\frac{λ}{R}\right)\hat{y} [/tex]

Thank you for the help.