Electric Field Intensity. Finite sheet of charge.

[azizlwl]

problem statement, all variables and given/known data

A finite sheet of charge, of density ρ=2x(x²+y²+4)^3/2, lies in the z=0 plane for 0≤x≤2m and 0≤y≤2m.Determine E at (0,0,2)m

Ans:(18x10^9)(-16/3a_x-4a_y+8a_z)

Homework Equations

E=kQ/R²

The Attempt at a Solution

dE=ρdA / R^2 a_R
dA=dxdy[/B]

E=k∫∫ 2x(x²+y²+4)^3/2 dy dx (-xa_x-ya_y+2a_z ) /x²+y²+4)^3/2

E=k∫∫ 2x dy dx (-xa_x-ya_y+2a<sub>z)

E=(18x10^9)(-4ax-4ay+8az)

How x and y component of E are different in the answer?</sub>

 

[BvU]

Do you see a difference between the x- and the y-dependence of $\rho$ ?

Or: Can you see x = y is not a symmetry line ?

(edit: forgot the +4 in the first upload)

Once again: making a drawing is soooooo useful !

 

[azizlwl]

Thanks for the visual. Is there method to find a point of symmetry on that surface?

 

[BvU]

What if there isn't ?

 

[azizlwl]

What if there isn't ?

Do I have to use mathematical theorem? If so I have to give up on this problem. I know how to use Coulomb law and multiple integration with help of Mathematica. This question no 2.50 from Schaum's Electromagnetics.

 

[BvU]

Ah, didn't really read the small print and just answered the question "How x and y component of E are different in the answer?". So now it's time to go through your steps and see why you and Schaum end up with different answers.
Let me please use $\hat\imath,\ \hat \jmath ,$ and $\hat k$ instead of $a_x, a_y, a_z$ as Cartesian unit vectors.

Stumble on the first step:
Your relevant equation only describes $|\vec E|\$, so it doesn't help you find the components.
Google (or look up in Schaum) the right equation and look at it carefully: the only vector that determines the direction of $|\vec E|\$ in there is $\vec r-\vec r'$ and you can't take it out of the integral because it varies with $d\vec r' = dx dy$.

Attempt at solution starts with

E=k∫∫ 2x(x2+y2+4)^3/2 dy dx (-xax-yay+2az ) /x2+y2+4)^3/2

which does not match with

dE=ρdA / R² a_R

(it looks to me that even the brackets don't match...)

My advice: Don't give up. This exercise is very important for your understanding and further curriculum. Make a drawing that shows $\ \vec E\$ at $(0,0,2)$ due to a small patch of charge $\rho dxdy$ located at $\vec r' = (x, y)$ in the x,y plane. You will see that the maker of the exercise chose a very nice expression for $\rho$.

The consider what you have to integrate to find the x-component of $\vec E$, $\ \vec E_x = \vec E \cdot \hat\imath \$, and remember that you can bring constant factors into the integral. Then idem the y component and the z-component.

Sounds perhaps a bit cryptic. But you should really work this out yourself - if I do it for you it does not help you. So try a first step and post when you get stuck.

 

[BvU]

Nice of you to like my post. I hope it's for the encouragement, because I am far too critical about your contents. Let me repair the expression with an extra bracket and look at it again...

 

[BvU]

OK, so it's all pretty minor. You're doing great and the only glitch is for the x component. All my well-intended teaching blabla doesn't help if you do the $Ey = k\iint 2xy\, dx dy$ completely correctly and all you are missing is the $Ex = k\iint 2x^2 \,dx dy$. Look at this last one again and tell me it was in fact easy ! (and hopefully also that you could have done it without crutches like Mathematica (*) just as well !)

I did find a worked out example for the first few steps. Later on he brings $\sigma = \rho$ outside the integral because it's constant -- you can't do that, but thanks to the nice $\rho$ function you could do other things to get simple integrals...and you did. It was just the last step that went off the rails a little bit.

(*) I have nothing against Mathematica, but here it's just like using a calculator for 2 x 3

 

[azizlwl]

Thanks. My fault is on the unit vector. I used to deal with constant z component.
So the unit vector should be (- xi - yj+2k)/|R|. Maybe using a_a, a_y and a_z is confusing.