Electric field of 3 quarters of an annulus

[fluidistic]

Homework Statement

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".

Homework Equations

None given.
d \vec E =\frac{kdQ \vec r}{r^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?

I thank you in advance.

 

[ehild]

Show your work more detailed. I got different result.

ehild

 

[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 )<br />. Though now I don't have much confidence in it.
Do you see any error? Do you want me to show more details?

 

[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 E_y 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

 

[fluidistic]

Ok thank you very much. That was really helpful!

 

[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.

 

[ehild]

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.

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

 

[fluidistic]

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

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.

 

[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.

 

[ehild]

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


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

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

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



ehild

 

[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.