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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: 20,939 Thanks: 2210 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: 2,258 Thanks: 929 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: 2,258 Thanks: 929 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 Tags divergence, field Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post harley05 Calculus 3 June 10th, 2014 10:16 AM rayman Abstract Algebra 1 March 11th, 2014 03:27 PM solrob Applied Math 2 November 10th, 2013 09:06 AM watson Abstract Algebra 1 September 14th, 2012 09:07 PM MasterOfDisaster Calculus 2 September 26th, 2011 09:17 AM

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