# Semi-circle linear charge Electric Field

1. Aug 28, 2016

### RJLiberator

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

View attachment 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.

$$E = \left(\frac{1}{4πε_0}\right)\left(\frac{dq}{R^2}\right)sin(θ)$$

Also: $$\left(\frac{dq}{dθ}\right) = \left(\frac{Q}{π}\right)$$

so: $$dq = \left(\frac{Qdθ}{π}\right)$$

Now:

$$E = \left(\frac{1}{4πε_0}\right)\left(\frac{Qsin(θ)}{πR^2}\right)dθ$$

After calculating the integral from θ = 0 to θ = π I get the following answer:

$$E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^2}\right)\hat{y}$$

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:

$$E = \left(\frac{1}{2πε_0}\right)\left(\frac{λ}{R^2}\right)\hat{y}$$

2. Aug 28, 2016

### blue_leaf77

I cannot open the file.
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.

Last edited by a moderator: May 8, 2017
3. Aug 28, 2016

### SammyS

Staff Emeritus
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.)

4. Aug 29, 2016

### 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:

$$E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^3}\right)\hat{y}$$

5. Aug 29, 2016

### blue_leaf77

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

6. Aug 29, 2016

### RJLiberator

If linear charge density
$$λ = \left(\frac{Q}{πR}\right)$$

$$E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^2}\right)\hat{y}$$
$$E = \left(\frac{1}{2πε_0}\right)\left(\frac{λ}{R}\right)\hat{y}$$