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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 3.38

Let's start by recalling Eq. 3.103 as found on section 3.4.4: The Electric Field of a Dipole:

$$\begin{array}{}\text{(Eq. 3.103, page 155)}& {\mathbf{E}}_{\text{dip}}(r,\theta )=\frac{p}{4\pi {ϵ}_0{r}^3}(2\cos \theta \hat{\mathbf{r}}+\sin \theta \hat{\mathit{\theta }})\end{array}$$

There are a few ways to jump from 3.103 to 3.104 and vice versa. My favorite is through the definition of dot products as projections. Any vector, say $\mathbf{a}=(a_x,a_y)$, can be written as:

$$\mathbf{a}=a_x\hat{\mathbf{x}}+a_y\hat{\mathbf{y}}$$

So far this is just notation. Now, what is $a_x$? It is the $x$-component of $\mathbf{a}$. We find this component through a projection:

$$a_x=\mathbf{a}\cdot \hat{\mathbf{x}}$$

In this sense, we write:

$$\mathbf{p}=p_r\hat{\mathbf{r}}+p_{\theta }\hat{\mathit{\theta }}=(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}+(\mathbf{p}\cdot \hat{\mathit{\theta }})\hat{\mathit{\theta }}$$

Replace $\mathbf{p}$ in Eq. 3.104:

$$\begin{array}{rl}{\mathbf{E}}_{\text{dip}}(\mathbf{r})& =\frac{1}{4\pi {ϵ}_0}\frac{1}{r^3}[3(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-(\mathbf{p}\cdot \hat{\mathit{\theta }})\hat{\mathit{\theta }}]\\ & =\frac{1}{4\pi {ϵ}_0}\frac{1}{r^3}[2(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-(\mathbf{p}\cdot \hat{\mathit{\theta }})\hat{\mathit{\theta }}]\end{array}$$

We now use the fact

$$\begin{array}{rl}\mathbf{p}\cdot \hat{\mathbf{r}}& =p\cos \left(\theta \right)\\ \mathbf{p}\cdot \hat{\mathit{\theta }}& =-p\sin \left(\theta \right)\end{array}$$

And substitute above:

$$\begin{array}{rl}{\mathbf{E}}_{\text{dip}}(\mathbf{r})& =\frac{1}{4\pi {ϵ}_0}\frac{1}{r^3}[2(\mathbf{p}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-(\mathbf{p}\cdot \hat{\mathit{\theta }})\hat{\mathit{\theta }}]\\ & =\frac{1}{4\pi {ϵ}_0}\frac{1}{r^3}[2(p\cos \left(\theta \right))\hat{\mathbf{r}}-(-p\sin \left(\theta \right))\hat{\mathit{\theta }}]\\ & =\frac{1}{4\pi {ϵ}_0}\frac{1}{r^3}[2(p\cos \left(\theta \right))\hat{\mathbf{r}}+(p\sin \left(\theta \right))\hat{\mathit{\theta }}]\\ & =\frac{p}{4\pi {ϵ}_0}\frac{1}{r^3}[2(\cos \left(\theta \right))\hat{\mathbf{r}}+(\sin \left(\theta \right))\hat{\mathit{\theta }}]\end{array}$$

Which is precisely Eq. 3.103. If we want to go from $3.103$ to $3.104$, we can use the same relations for $\mathbf{p}\cdot \hat{\mathbf{r}}$ and $\mathbf{p}\cdot \hat{\mathit{\theta }}$ and build up from there.

If you are confused as to how to get the equations for $\mathbf{p}\cdot \hat{\mathbf{r}}$ and $\mathbf{p}\cdot \hat{\mathit{\theta }}$, consider the following:

The magnitude of $\hat{\mathbf{r}}$ and $\hat{\mathit{\theta }}$ is 1, of course. Further, notice how $\hat{\mathit{\theta }}$ tends to point opposite to $\mathbf{p}$, that's why the ($-$) sign (the dashed green vector is just $\hat{\mathit{\theta }}$ starting at the origin rather at its familiar location (at the tip of $\hat{\mathbf{r}}$)).

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