Calculate Nabla Operator for Potential Function with Distance r

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Does my solution look correct to you guys?

Homework Statement



Calculate:
\nabla \varphi (r)

If:
\varphi (r) = \frac{1}{4\pi\epsilon_{0}}\frac{1}{r}

with: r = \sqrt{x^{2}+y^{2}+z^{2}}

Homework Equations



n/a

The Attempt at a Solution



hw1.jpg

hw2.jpg
 
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You used the divergence not the gradient. Remember taking the gradient of a scalar field "creates" a vector.
 
Is this better? I can't figure out how to simplify it further.

hw4.jpg
 
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That is correct, but you can simplify it much further. What is x^2+y^2+z^2? What is x \hat{e_1}+y\hat{e_2}+z \hat{e_3}?

Hint: \vec{r}=?
 
It is the derivative w.r.t. r times r-hat, because if you change r by dr, the step length in the r-hat direction is dr.


By definition, you have for a differentiable function f(x1,x2,...) that:

df = nabla f dot ds (1)

where ds is the displacement vector.

You also have that:

df = df/dx1 dx1 + df/dx2 dx2 + df/dx3 dx3 + ... (2)

In case of Cartesian coordinates, ds = (dx1, dx2,...), but in general this is not the case.
 
Cyosis said:
What is x^2+y^2+z^2

r^{2} = x^{2}+y^{2}+z^{2}?

Cyosis said:
What is x \hat{e_1}+y\hat{e_2}+z \hat{e_3}
- \hat{r} = -x \hat{e_1}-y\hat{e_2}-z \hat{e_3}?


So final answer:
\frac{-\hat{r}}{r^{5/2}} ?
 
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No, |\hat{r}|=1 yet we know that |x \hat{e_1}+y\hat{e_2}+z \hat{e_3}| \neq 1. Again, do you know what \vec{r} looks like in Cartesian coordinates?
 
I'm thinking hard but I'm not sure..

I'm thinking something along the lines of.. the x-component of \hat{r}.. the y-component. or something? >_<
 
You're getting close. We are given that |\vec{r}|=\sqrt{x^2+y^2+z^2}=r. So we want to find s an expression for \vec{r} whose length is r and points in the \hat{r} direction.
 
  • #10
I don't understand the differences between:

\vec{r},\hat{r} and r!

=(
 
  • #11
A vector has a direction and a magnitude. A unit vector is a vector with magnitude 1. For example \hat{e_1} is the unit vector pointing in the x-direction with |\hat{e_1}|=1. These are used in Cartesian coordinates and I am sure you're familiar with them. In your problem however we are working with spherical coordinates. the unit vectors for spherical coordinates are \hat{r},\hat{\theta}, \hat{\phi}. So whereas you express a vector in Cartesian coordinates as a \hat{e_1}+b\hat{e_2}+c \hat{e_3} in spherical coordinates you would express it as \alpha \hat{r}+\beta \hat{\theta}+\gamma \hat{\phi}.

In spherical coordinates the length of a vector \vec{r} is given by |\vec{r}|=\sqrt{x^2+y^2+z^2}=r. Therefore r is the magnitude of \vec{r}. A vector also has a direction and the direction of \vec{r} is given by\hat{r}.

It is given in the problem statement that |\vec{r}|=\sqrt{x^2+y^2+z^2} and we have a vector,x \hat{e_1}+y\hat{e_2}+z \hat{e_3} in Cartesian coordinate which we would like to write as a vector in spherical coordinates. If you draw a Cartesian coordinate system and draw a radial vector on the axes, then how can you express this radial vector in terms of x,y,z? Compare it to the vector we're trying to write in spherical coordinates, also compare the length of x \hat{e_1}+y\hat{e_2}+z \hat{e_3} with the length of \vec{r}.

Lastly you also need to write (x^2+y^2+z^2)^{-3/2} in terms of r correctly, it is not r^{-5/2}.
 
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  • #12
bumclouds said:
So final answer:
\frac{-\hat{r}}{r^{5/2}} ?

This is almost there... if you fix your confusion in how to notate each kind of vector, you only have to fix a minor arithmetical error
 
  • #13
=\frac{-r\hat{r}}{r^{3}}

??

so it's the negative r unit vector pointing in the r direction.. over r cubed.. right? =(

I seem to have a huge knowledge hole about vectors
 
  • #14
It is the correct answer, but how did you get to it?

so it's the negative r unit vector pointing in the r direction.. over r cubed.. right?

No it is the unit vector pointing in the -\hat{r} direction multiplied by the scalar r. That means it is a vector pointing radially towards the origin of the coordinate system and has a length r.
 
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