Deriving electric field with given potential

[AndrewAmmerlaa]

Homework Statement

Given that the potential of dipole is equal to:
$$V(\vec{r})=\frac{\vec{p}\vec{r}}{4\pi\epsilon_0 r^3}$$
show that the electric field is equal to:
$$\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}(\frac{3(\vec{p}\vec{r})\vec{r}}{r^5}-\frac{\vec{p}}{r^3})$$

Homework Equations

$$\vec{E}(\vec{r})=-\nabla V(\vec{r})$$

The Attempt at a Solution

I thought i could just differentiate V(r):

$$\vec{E}(\vec{r})=-\nabla V(\vec{r}) = \frac{-1}{4\pi\epsilon_0}(\frac{r^3\vec{p}-3r^2\vec{p}\vec{r}}{r^6})\hat{r}=\frac{1}{4\pi\epsilon_0}(\frac{3(\vec{p}\vec{r})\vec{r}}{r^5}-\frac{\vec{p}\vec{r}}{r^4})$$
but that's not the correct answer, my question is: why does only one of the terms get multiplied with r^hat? It seems to me that the nabla operator would multiply the whole thing with r^hat, but that's not the correct answer according to the question (I checked if the question is correct and it is)

 

[drvrm]

but that's not the correct answer, my question is: why does only one of the terms get multiplied with r^hat? It seems to me that the nabla operator would multiply the whole thing with r^hat, but that's not the correct answer according to the question (I checked if the question is correct and it is)[/code]

use correct form of Nabla operator .

V(⃗r)=⃗p⃗r4πϵ0r3V(r→)=p→r→4πϵ0r3

in your expression for potential there is there should be p.r .

 

[AndrewAmmerlaa]

use correct form of Nabla operator .

I used this, is this not the correct form?
$$\nabla f=\frac{\partial f}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial f}{\partial \theta}\hat{\theta}+\frac{1}{rsin{\theta}}\frac{\partial f}{\partial \phi}\hat{\phi}$$
V is not dependent on phi or theta so the derivative of that is zero, that just leaves that partial derivative to r times r^hat

 

[drvrm]

dipole potential has a standard form -one can see in any textbook

V(r,theta)= p.r/ 4Pi.epsilon.r^2 so check in the question !

 

[AndrewAmmerlaa]

dipole potential has a standard form -one can see in any textbook

V(r,theta)= p.r/ 4Pi.epsilon.r^2 so check in the question !

I don't understand, according to wikipedia the equation for potential is correct: https://en.wikipedia.org/wiki/Dipole#Field_from_an_electric_dipole

 

[ehild]

I used this, is this not the correct form?
$$\nabla f=\frac{\partial f}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial f}{\partial \theta}\hat{\theta}+\frac{1}{rsin{\theta}}\frac{\partial f}{\partial \phi}\hat{\phi}$$
V is not dependent on phi or theta so the derivative of that is zero, that just leaves that partial derivative to r times r^hat

V can depend on phi and theta through the product of the position vector $\vec r$ with the vector $\vec p$. For example, if $\vec p$ is parallel with the z axis, $\vec p\cdot \vec r = pr\cos(\theta)$Only a spherically symmetric potential depends only on r.
Use the form of the nabla operator in Cartesian coordinates. What is it?
If you check your result for the electric field you can see that one term is a vector, the other is a scalar, which is certainly wrong.

 

[AndrewAmmerlaa]

V can depend on phi and theta through the position vector $\vec r$. Only a spherically symmetric potential depends only on r.
Use the form of the nabla operator in Cartesian coordinates. What is it?
If you check your result for the electric field you can see that one term is a vector, the other is a scalar, which is certainly wrong.

I hadn't thought of that, adding the derivative of V to theta gives an term with $\hat{\theta}$ and i have no idea what to do with that.

nabla in cartesian coordinates:
$$\nabla f= \frac{\partial f}{\partial x}\hat{x}+\frac{\partial f}{\partial y}\hat{y}+\frac{\partial f}{\partial z}\hat{z}$$

but that gives me this:
$$\vec{E}(\vec{r})=-\nabla \frac{\vec{p}\vec{r}}{4\pi\epsilon_0 r^3}=-\frac{r^3\vec{p}-3r^2\vec{p}\vec{r}}{r^6}\hat{x}-\frac{r^3\vec{p}-3r^2\vec{p}\vec{r}}{r^6}\hat{y}-\frac{r^3\vec{p}-3r^2\vec{p}\vec{r}}{r^6}\hat{z}=\frac{1}{4\pi\epsilon_0}(\frac{3\vec{p}\vec{r}}{r^4}-\frac{\vec{p}}{r^3})(\hat{x}+\hat{y}+\hat{z})$$ but that leaves me one $\vec r$ short in the first term, and $\hat{x}+\hat{y}+\hat{z}$ would be the vector (1,1,1) and I'm unsure what to do with that.

 

[ehild]

I hadn't thought of that, adding the derivative of V to theta gives an term with $\hat{\theta}$ and i have no idea what to do with that.

nabla in cartesian coordinates:
$$\nabla f= \frac{\partial f}{\partial x}\hat{x}+\frac{\partial f}{\partial y}\hat{y}+\frac{\partial f}{\partial z}\hat{z}$$

but that gives me this:
$$\vec{E}(\vec{r})=-\nabla \frac{\vec{p}\vec{r}}{4\pi\epsilon_0 r^3}=-\frac{r^3\vec{p}-3r^2\vec{p}\vec{r}}{r^6}\hat{x}-\frac{r^3\vec{p}-3r^2\vec{p}\vec{r}}{r^6}\hat{y}-\frac{r^3\vec{p}-3r^2\vec{p}\vec{r}}{r^6}\hat{z}=\frac{1}{4\pi\epsilon_0}(\frac{3\vec{p}\vec{r}}{r^4}-\frac{\vec{p}}{r^3})(\hat{x}+\hat{y}+\hat{z})$$ but that leaves me one $\vec r$ short in the first term, and $\hat{x}+\hat{y}+\hat{z}$ would be the vector (1,1,1) and I'm unsure what to do with that.

That is wrong. Write out the whole potential in Cartesian coordinates, and take the partial derivatives with respect x, y, z.

 

[AndrewAmmerlaa]

Ok, so I got this:
$$V(\vec{r})=V(x,y,z)=\frac{1}{4\pi\epsilon_0(x^2+y^2+z^2)^{\frac{3}{2}}}\begin{pmatrix}p_1 \\ p_2 \\ p_3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}=\frac{p_1x+p_2y+p_3z}{4\pi\epsilon_0(x^2+y^2+z^2)^{\frac{3}{2}}} \implies \vec{E}(\vec{r})=\frac{-1}{4\pi\epsilon_0}(\frac{\sqrt{x^2+y^2+z^2}^3p_1-3x\sqrt{x^2+y^2+z^2}p_1x}{\sqrt{x^2+y^2+z^2}^6}\hat{x}$$
plus simular terms for y and z
$$=\frac{1}{4\pi\epsilon_0}(\frac{3(p_1x^2\hat{x}+p_2y^2\hat{y}+p_3z^2\hat{z})}{\sqrt{x^2+y^2+z^2}^5}-\frac{p_1\hat{x}+p_2\hat{y}+p_3\hat{z}}{\sqrt{x^2+y^2+z^2}^3}=\frac{1}{4\pi\epsilon_0}(\frac{3(p_1x^2\hat{x}+p_2y^2\hat{y}\hat{x}+p_3z^2\hat{z})}{r^5}-\frac{p_1\hat{x}+p_2\hat{y}+p_3\hat{z}}{r^3})$$
how do I turn the p1+p2+p3 back into a vector? I'm sorry I'm asking so many questions i thought i understood vector calculus but it turns out i don't.

EDIT: never mind, i forgot the hats, that makes adding p1 p2 and p3 easy, but what about the p1x^2+p2y^2+p3z^2?
wouldn't that be something like this:
$$\begin{pmatrix} p_1x^2\\ p_2y^2\\ p_3z^2\\ \end{pmatrix}$$ instead of $(\vec{p}\vec{r})\vec{r}$

 

[ehild]

Ok, so I got this:
$$V(\vec{r})=V(x,y,z)=\frac{1}{4\pi\epsilon_0(x^2+y^2+z^2)^{\frac{3}{2}}}\begin{pmatrix}p_1 \\ p_2 \\ p_3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}=\frac{p_1x+p_2y+p_3z}{4\pi\epsilon_0(x^2+y^2+z^2)^{\frac{3}{2}}} \implies \vec{E}(\vec{r})=\frac{-1}{4\pi\epsilon_0}(\frac{\sqrt{x^2+y^2+z^2}^3p_1-3x\sqrt{x^2+y^2+z^2}p_1x}{\sqrt{x^2+y^2+z^2}^6}\hat{x}$$

You made a mistake. When you take the derivative of the denominator, you have to keep the whole numerator.

 

[AndrewAmmerlaa]

I got it:
$$V(\vec{r})=V(x,y,z)=\frac{1}{4\pi\epsilon_0(x^2+y^2+z^2)^{\frac{3}{2}}}\begin{pmatrix}p_1 \\ p_2 \\ p_3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}=\frac{p_1x+p_2y+p_3z}{4\pi\epsilon_0(x^2+y^2+z^2)^{\frac{3}{2}}} \implies \vec{E}(\vec{r})=\frac{-1}{4\pi\epsilon_0}(\frac{\sqrt{x^2+y^2+z^2}^3p_1-3x\sqrt{x^2+y^2+z^2}(p_1x+p_2y+p_3z)}{\sqrt{x^2+y^2+z^2}^6}\hat{x}$$
plus simular terms for y and z
$$=\frac{1}{4\pi\epsilon_0}(\frac{3x(p_1x+p_2y+p_3z)\hat{x}+3y(p_1x+p_2y+p_3z)\hat{y}+3z(p_1x+p_2y+p_3z)\hat{z}}{\sqrt{x^2+y^2+z^2}^5}-\frac{p_1\hat{x}+p_2\hat{y}+p_3\hat{z}}{\sqrt{x^2+y^2+z^2}^3}=\frac{1}{4\pi\epsilon_0}(\frac{3x\vec{p}\vec{r}\hat{x}+3y\vec{p}\vec{r}\hat{y}+3z\vec{p}\vec{r}\hat{z}}{r^5}-\frac{p_1\hat{x}+p_2\hat{y}+p_3\hat{z}}{r^3})= \frac{1}{4\pi\epsilon_0}(\frac{3\vec{p}\vec{r}\vec{r}}{r^5}-\frac{\vec{p}}{r^3})$$

Thanks so much, this was a tough one.

 

[ehild]

I got it:
$$= \frac{1}{4\pi\epsilon_0}(\frac{3\vec{p}\vec{r}\vec{r}}{r^5}-\frac{\vec{p}}{r^3})$$

Correctly:
$$= \frac{1}{4\pi\epsilon_0}(\frac{3(\vec{p}\vec{r})\vec{r}}{r^5}-\frac{\vec{p}}{r^3})$$

There is a simpler way to find the gradient: It can be considered as derivative with respect $\vec r$.
r is the magnitude of $\vec r$ so $r=\sqrt{{\vec r}^2}$
$$V(\vec{r})=\frac{\vec p \cdot \vec r }{4\pi\epsilon_0 r^3}=\frac{1}{4\pi\epsilon_0}\frac{\vec p \cdot \vec r }{\left( {\vec r}^2\right)^{3/2}}$$
$$\vec E = - \frac{dV}{d \vec r }=- \frac{1}{4\pi\epsilon_0}\frac{\vec p \left( {\vec r}^2\right)^{3/2} -(\vec p \cdot \vec r)3(\vec r^2)^{1/2} \vec r}{\left( {\vec r}^2\right)^{3/2}}=-\frac{1}{4\pi\epsilon_0}\frac{\vec p r^3 -3(\vec p \cdot \vec r)r \vec r}{r^6}$$