Potential and Field

Question

So given a field:

$$\vec E(r)=\frac{\alpha(\vec p \cdot \vec e_r)\vec e_r + \beta \vec p}{r^3}$$

where $α, β$ are constants, $\vec e_r$ is the unit vector in the direction $\vec r$, and $\vec p$ is a constant vector.

I'm supposed to find out the relationship between $\alpha$ and $\beta$ such that $\vec E$ is a physical field.

My first instinct is, since the field s given in direction of $r$, I'm going to use spherical coordinates to make my life easier and then calculate $\nabla \times E = 0$.

I know that in spherical coordinates

$$\nabla \times E = \frac{1}{r\sin \theta} \Big\{ \frac{\partial}{\partial \theta}(\sin \theta V_{\theta}) - \frac{\partial V_{\theta}}{\partial \theta}\Big\} + \cdots$$

For obvious reasons, I'm not typing everything out.

My first question is, when we look at the field

$$\vec E(r)=\frac{\alpha(\vec p \cdot \vec e_r)\vec e_r + \beta \vec p}{r^3}$$

isn't this the same as

$$\vec E(r)=\frac{\alpha\vec p \cdot \vec e_r + \beta \vec p}{r^3}$$

because a unit vector dotted with itself is $1$?

My second question: the direction of the $P$ vector isn't given except in the first part, how do I relate it to $\theta$ head and $\phi$ head?

Third question (sorry for my shaky calculus foundation): taking the example of the first term of the curl of the field (which I typed out in LaTeX), since nothing in the $E$ field has a $\theta$ or $\phi$ term, the first term is $0$? I guess my insecurity about this assumption is because we have a constant vector pointing in some direction which unless is completely parallel to $r$ head we can relate its direction to $r$ head with appropriate $\theta$ and $\phi$ terms?

Thanks for any help!

Answer 1

For the first question, $\vec p\cdot\vec e_r$ is a scalar but $(\vec p\cdot\vec e_r)\vec e_r$ is a vector by elementwise multiplication, so you cannot simplify like that.

For the second question, there is no need to use explicit $p_r,p_\theta,p_\varphi$ components – without loss of generality set $\vec p$ in the same direction as $\theta=0$ (i.e. the $+z$-axis). Then (third question) $\vec p=p(\cos\theta,-\sin\theta,0)^T$, its dot product with $\vec e_r=(1,0,0)^T$ is $p\cos\theta$ and $$\vec E=\frac p{r^3}\begin{bmatrix} (\alpha+\beta)\cos\theta\\ -\beta\sin\theta\\ 0\end{bmatrix}$$ The $\varphi$-component being zero, only that component of the curl will be nonzero and it equals $$\frac{(rE_\theta)_r-(E_r)_\theta}r=p\cdot\frac{(2\beta\sin\theta)/r^3+((\alpha+\beta)\sin\theta)/r^3}r=\frac{p(\alpha+3\beta)\sin\theta}{r^4}$$ Thus $\vec E$ is irrotational iff $\alpha+3\beta=0$, in which case $$\vec E=-\frac{p\beta}{r^3}\begin{bmatrix} 2\cos\theta\\ \sin\theta\\ 0\end{bmatrix}=\nabla\frac{p\beta\cos\theta}{r^2}$$

Answer 2

Because the expression for the field only contains $\hat{\mathbf{e}}_r$, it does not pin down a direction for $\theta = 0$, so we can choose it to be whatever is most convenient. In this case, the simplest choice is to be parallel to $\mathbf{p}$, the only preferred directon in the problem setup. Thus we will have $\mathbf{p} = p \hat{\mathbf{e}}_p$, where $\hat{\mathbf{p}}$ is the direction of $\theta = 0$ (in cylindrical coordinates, this would be $\hat{\mathbf{e}}_z$). This unit vector can be expressed as $\hat{\mathbf{e}}_p = \cos\theta \hat{\mathbf{e}}_r -\sin\theta \hat{\mathbf{e}}_\theta$. If you don't see why, try converting it to cylindrical coordinates. You'll find that you get $\hat{\mathbf{e}}_z$ as expected. This means $\hat{\mathbf{e}}_r\cdot\mathbf{p} = p\cos\theta$, and we have \begin{eqnarray} \mathbf{E}(\mathbf{r}) &=& \frac{\alpha p\cos\theta \hat{\mathbf{e}}_r + \beta p (\cos\theta \hat{\mathbf{e}}_r - \sin\theta\hat{\mathbf{e}}_\theta)}{r^3} \\&=& \frac{p}{r^3}\left[(\alpha + \beta)\cos\theta \hat{\mathbf{e}}_r - \beta\sin\theta \hat{\mathbf{e}}_\theta\right] \end{eqnarray} This doesn't depend on $\phi$ or have any $\hat{\mathbf{e}}_\phi$ component, so only the $\hat{\mathbf{e}}_\phi$ component of the curl can be nonzero. We can evaluate that as \begin{eqnarray} (\nabla\times\mathbf{E})_\phi &=& \frac{1}{r}\left[-\beta p\frac{\partial}{\partial r}\left(r\frac{ \sin\theta}{r^3}\right) - (\alpha + \beta)p\frac{\partial}{\partial\theta}\left(\frac{\cos\theta}{r^3}\right)\right] \\&=& \frac{ p\sin\theta}{r^4}(\alpha +3\beta), \end{eqnarray} which means $\nabla\times\mathbf{E} = 0$ if $\alpha = -3\beta$, and the full expression for $\mathbf{E}$ is $$ \mathbf{E}(\mathbf{r}) = -\frac{p\beta}{r^3}\left[2\cos\theta \hat{\mathbf{e}}_r + \sin\theta \hat{\mathbf{e}}_\theta\right]. $$ It's not too difficult from here to figure out what function $V$ gives $-\nabla V = \mathbf{E}$, so I'll leave that to you.