My Math Forum Divergence of Field

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 March 29th, 2019, 10:54 AM #1 Newbie   Joined: Mar 2019 From: a Posts: 3 Thanks: 1 Divergence of Field Hi. I got the following field: $\displaystyle E = \frac{ kQ \cdot (r\vec - R/2 x \hat )}{/(8 (r(vec) - R/2 x(hat))^3}$ I'm getting trouble to do a divergence to this field. I'm not sure in what I've did but I wrote this expression as: E= kQ/(8(r^2 - (R/2)^2)) and than did a divergence for it.. but it seems wrong expression. by the way, how can I write in this forum in LaTeX code or something? thanks. Last edited by 1993kosta; March 29th, 2019 at 11:13 AM.
March 29th, 2019, 11:50 AM   #2
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 Originally Posted by 1993kosta Hi. I got the following field: $\displaystyle E = \frac{ kQ \cdot (r\vec - R/2 x \hat )}{/(8 (r(vec) - R/2 x(hat))^3}$ I'm getting trouble to do a divergence to this field. I'm not sure in what I've did but I wrote this expression as: E= kQ/(8(r^2 - (R/2)^2)) and than did a divergence for it.. but it seems wrong expression. by the way, how can I write in this forum in LaTeX code or something? thanks.
First off you have to keep the vectors when doing the divergence. For example, the divergence of $\displaystyle \vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k}$

is (in rectangular coordinates) $\displaystyle \vec{ \nabla } \cdot \vec{E} = \dfrac{\partial E_x}{\partial x} + \dfrac{ \partial E_y }{\partial y} + \dfrac{ \partial E_z }{ \partial z}$

Yes, you have some LaTeX problems. Based on your magnitude for $\displaystyle \vec{E}$ I would guess that it should be
$\displaystyle \vec{E} = \dfrac{kQ~ \left ( \vec{r} - \dfrac{1}{2} \hat{R} \right )}{\left | \left ( 8\vec{r} - \dfrac{1}{2} \hat{R} \right ) \right | ^3}$

That's not to say I have no doubts. I am at a loss to explain the 8 and I don't know why you would have $\displaystyle \hat{R}$ instead of $\displaystyle \vec{R}$, so please verify that I have the correct form.

-Dan

 March 29th, 2019, 03:00 PM #3 Newbie   Joined: Mar 2019 From: a Posts: 3 Thanks: 1 That's the right one: $\displaystyle \vec{E} = \dfrac{kQ ( \vec{r} - \dfrac{1}{2} R\hat{x} )}{8 | ( \vec{r} - \dfrac{1}{2} R\hat{x}) | ^3}$ I will try to explain with words because it doesn't work for me well. The expression is with x(hat) instead of R hat (there is a r hat. Plus the '8' is outside the expression: 8|r(vec)-(R/2)x(hat). In this problem, should I do divergence in spherical or Cartesian coordinates? (because I have r and x) Does this expression equal $\displaystyle \vec{E} = \dfrac{kQ~ }{8\left| \left( \vec{r} - \dfrac{1}{2} R\hat{x} \right) \right| ^2}$ Thanks Thanks from topsquark Last edited by skipjack; March 29th, 2019 at 06:28 PM.
March 29th, 2019, 06:28 PM   #4
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 Originally Posted by 1993kosta That's the right one: $\displaystyle \vec{E} = \dfrac{kQ ( \vec{r} - \dfrac{1}{2} R\hat{x} )}{8 | ( \vec{r} - \dfrac{1}{2} R\hat{x}) | ^3}$ I will try to explain with words because It doesn't work for me well. The expression is with x(hat) instead of R hat (there is a r hat. Plus the '8' is outside the expression: 8|r(vec)-(R/2)x(hat). In this problem, should I do divergence in spherical or Cartesian coordinates? (because I have r and x) Does this expression equal $\displaystyle \vec{E} = \dfrac{kQ~ }{8\left| \left( \vec{r} - \dfrac{1}{2} R\hat{x} \right) \right| ^2}$ Thanks
My suggestion would be to convert the $\displaystyle \hat{x}$ to spherical coordinates.

Also, your suggested cancellation is not correct as the RHS is not a vector!

Let's define $\displaystyle \vec{T} = \vec{r} - \dfrac{1}{2} R \hat{x}$. (In order to keep the notation simpler.)

$\displaystyle \vec{E} = \dfrac{kQ ( \vec{r} - \dfrac{1}{2} R\hat{x} )}{8 | ( \vec{r} - \dfrac{1}{2} R\hat{x}) | ^3} = \dfrac{kQ \vec{T}}{8 |\vec{T} |^3} = \dfrac{kQ \hat{T}}{8 |\vec{T} |^2}$
Where T is the unit vector along $\displaystyle \vec{T}$. (We can also say that $\displaystyle |\vec{T}|^2$ is simply $\displaystyle \vec{T} \cdot \vec{T} = T^2$.)

-Dan

Last edited by topsquark; March 29th, 2019 at 06:36 PM.

 March 29th, 2019, 06:44 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,485 Thanks: 2041 The second isn't a vector. Thanks from topsquark and 1993kosta
March 29th, 2019, 10:37 PM   #6
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ok, so now can I do the divergence on vector T?
How can I know the nabla for this divergence?

thanks, kosta.

Edit:
Is what I did in the picture is right?
Attached Images
 pic.jpg (81.5 KB, 2 views)

Last edited by 1993kosta; March 29th, 2019 at 11:23 PM.

 March 30th, 2019, 09:53 AM #7 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timey-wimey stuff. It's a bit more complicated that that. That $\displaystyle \hat{x}$ is a rather large problem. But in any case $\displaystyle \hat{x} = \sin( \theta ) ~ \cos( \phi ) ~ \hat{r} + r\cos( \theta ) ~ \cos( \phi ) ~ \hat{ \theta }- r\sin( \theta ) ~ \sin( \phi ) ~ \hat{ \phi }$ (If I remember it correctly, anyway. It's been a little while.) So you have $\displaystyle \vec{E} = kQ \frac{ \vec{r} - \frac{R}{2} ( \sin( \theta ) ~ \cos( \phi ) ~ \hat{r} + r\cos( \theta ) ~ \cos( \phi ) ~ \hat{ \theta }- r\sin( \theta ) ~ \sin( \phi ) ~ \hat{ \phi } ) }{ 8 \left | \vec{r} - \frac{R}{2}( \sin( \theta ) ~ \cos( \phi ) ~ \hat{r} + r\cos( \theta ) ~ \cos( \phi ) ~ \hat{ \theta }- r\sin( \theta ) ~ \sin( \phi ) ~ \hat{ \phi } ) \right | ^3 }$ There's another mistake in your work. This is the second time this has come up so please make sure you understand it. Your form for $\displaystyle \vec{E}$ is still wrong. Once again you have neglected to consider that $\displaystyle \vec{E}$ is a vector. When you did your cancellation in this last post you, once again, tried to cancel out the vector in the numerator. The numerator is a vector. You need to keep this in mind. You know, this is a nastier problem than I had originally thought. It doesn't look any nicer to use rectangular coordinates instead. Where did you get this problem? -Dan Last edited by skipjack; March 30th, 2019 at 10:38 AM.
 March 30th, 2019, 02:57 PM #8 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timey-wimey stuff. Actually it's not so bad after all. If you expand the vector $\displaystyle \vec{r}$ into its components then there is a lot of simplification you can do. So give that a try and if you can't figure it out just let us know. -Dan

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