# Electric field of 3 quarters of an annulus

1. Jan 29, 2010

### fluidistic

1. The problem statement, all variables and given/known data
There's 3 quarters of an annulus with inner radius a and outer radius b. It occupies the first 3 quadrants. Its charge density is $$\sigma$$. I must calculate the electric field at the point in the middle of the "annulus".

2. Relevant equations None given.
$$d \vec E =\frac{kdQ \vec r}{r^3}$$.

3. The attempt at a solution
After 2 attempts, I came up with the idea of separating the annulus in this way: the left part of it, i.e. the part of the annulus situated in the 2nd and 3rd quadrants and the right part of it, i.e. the part situated in the first quadrant.
I calculated their contributions to the electric field to be $$k\pi \sigma \ln \left ( \frac{b}{a} \right ) \hat i$$ and $$(-\frac{\sqrt 2}{2}k\pi \sigma \ln \left ( \frac{b}{a} \right ) \hat i, -\frac{\sqrt 2}{2}k\pi \sigma \ln \left ( \frac{b}{a} \right ) \hat j)$$, respectively.
Adding them up, I finally reached $$\vec E = \left ( \left ( 1 - \frac{\sqrt 2}{2}\right ) k\pi \sigma \left ( \frac{b}{a} \right ) \hat i , -\frac{\sqrt 2}{2} k\pi \sigma \left ( \frac{b}{a} \right ) \hat j \right )$$.
Is my method for solving the problem good? Do you know any other way to solve it? Lastly, is my answer errorless?

2. Jan 30, 2010

### ehild

Show your work more detailed. I got different result.

ehild

3. Jan 30, 2010

### fluidistic

Thank you for taking time to solve the problem!
Ok, for the first separation, $$dE=\frac{k\pi r \sigma dr}{r^2}$$ because $$dQ=\pi r \sigma dr$$. Thus $$E=k \pi \sigma \int _a ^b \frac{dr}{r}$$, hence $$\vec E=k\pi \sigma \ln \left ( \frac{b}{a} \right )$$.

For the right part of the annulus, $$dA=\frac{\pi r dr}{2}\Rightarrow dQ=\frac{\pi r \sigma dr}{2}$$. From this, I get .... ah ok, I see a little error, now I get $$\vec E= \left ( -\frac{\sqrt 2}{4} k \pi \sigma \ln \left ( \frac{b}{a} \right ) \hat i , -\frac{\sqrt 2}{4} k \pi \sigma \ln \left ( \frac{b}{a} \right ) \hat j \right )$$.

So the total electric field at the point considered is $$\vec E = \left ( \left ( 1 - \frac{\sqrt 2}{4}\right ) k\pi \sigma \left ( \frac{b}{a} \right ) \hat i , -\frac{\sqrt 2}{4} k\pi \sigma \left ( \frac{b}{a} \right ) \hat j \right )$$. Though now I don't have much confidence in it.
Do you see any error? Do you want me to show more details?

4. Jan 30, 2010

### ehild

Do not forget that E is a vector and dE is parallel to r
If you have a dQ charge around the position (r,phi),

$$dQ =\sigma r d \varphi dr$$

its contribution to the x component of the electric field is

$$dE_x=-k\frac{dQ}{r^2}\cos{\varphi}$$

and the contribution to the y component of the field is

$$dE_y=-k\frac{dQ}{r^2}\sin{\varphi}$$

You have to integrate with respect to phi from 0 to 3/4 pi and with respect to r form a to b.

$$E_x=-k\int_a^b\int_0^{3\pi/4}{\frac{\sigma \cos{\varphi} dr d\varphi}{r}$$

$$E_y=-k\int_a^b\int_0^{3\pi/4}{\frac{\sigma \sin{\varphi} dr d\varphi}{r}$$

You used a very clever simplification: the left half will result in zero contribution to Ey and the third quadrant will contribute with equal amount to both Ex and Ey. But you need to get the same result as with the integration for the whole range of angles.

ehild

Last edited: Jun 29, 2010
5. Jan 30, 2010

### fluidistic

Ok thank you very much. That was really helpful!

6. Jan 31, 2010

### fluidistic

I'm still confused about why I don't get the result.
$$d\vec E$$ is indeed parallel to $$\vec r$$.
And so the formula $$d \vec E =\frac{kdQ \vec r}{r^3}$$ becomes $$dE =\frac{kdQ}{r^2}$$, right?
This is what I've done.

7. Jan 31, 2010

### ehild

dE is the magnitude of the vector dE. You have to add the contribution of each dQ to the electric field. These contributions are vectors.
The direction of each dE is parallel to that vector r which points to the actual dQ. You can not add the magnitude of vectors. You have to add their x components, to get the x component of the resultant and the same with the y components.

ehild

8. Jan 31, 2010

### fluidistic

Ah ok, I get you.
I will try it again via my way although yours is probably faster.
Thanks for all... and don't be surprised if I ask for further help!

Edit: I just realized my way is wrong and yours is right.

Last edited: Jan 31, 2010
9. Jan 31, 2010

### fluidistic

I reached $$\vec E = \left ( k\sigma \ln \left ( \frac{b}{a} \right ) \hat i, k\sigma \ln \left ( \frac{b}{a} \right ) \hat j \right )$$.
Can you confirm the answer? At least it makes some sense: the i and j components are the same, except for the sign, as expected.

10. Jan 31, 2010

### ehild

I made a mistake, the integrals go from 0 to 3pi/2.

$$E_x=-k\int_a^b\int_0^{3\pi/2}{\frac{\sigma \cos{\varphi} dr d\varphi}{r}$$

$$E_y=-k\int_a^b\int_0^{3\pi/2}{\frac{\sigma \sin{\varphi} dr d\varphi}{r}$$

Your result is OK but one typo, there should be a - sign in front of the y component.

ehild

11. Feb 1, 2010

### fluidistic

Ok thanks. Yes I made a typo! And I didn't "copy and paste" your work, so that I took the upper limit of integration as $$\frac{3 \pi}{2}$$.
Thank you very much for all.