<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/14e6a1b3-8a2b-42dc-b3a7-d0e5effc3019/dipole.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/14e6a1b3-8a2b-42dc-b3a7-d0e5effc3019/dipole.png" width="40px" /> A dipole, $p$, is formed from two equal and opposite charges ( $q$ and $-q$ ) that are separated by a displacement $\vec d$ such that

$$ \vec p=q\vec d $$

• bringing a charge near a insulator induces a temporary dipole

Example: water molecules have a permanent electric dipole

</aside>

Electric field in line with dipole

• We have an Electric field $\vec E$ emanating from a source $A$ at a distance $r$ above the center of the dipole. $\hat r$ is decided in the up direction and $\vec E$ is in the $-\hat r$ direction, towards the dipole.

$$ \vec E=\vec E_++\vec E_- = \frac{Q \, \hat r}{4\pi\epsilon_0 r_1^2}-\frac{Q \, \hat r}{4\pi\epsilon_0 r^2_2} $$

where $r_1=r-d/2$ and $r_2=r+d/2$

• Assuming $r\gg d$ we can use the simplification

$$ \frac{1}{r^2_1}=\frac{1}{\left(r-\frac{d}{2}\right)^2}=\frac{1}{r^2\left(1-\frac{d}{2r}\right)^2}\approx\frac{1+\frac{d}{r}}{r^2} $$

• Adding this to our equation for $\vec E$ we get

$$ \vec E=\frac{Q \, \hat r}{4\pi\epsilon_0 r^2}\left[ \left(1+\frac{d}{r} \right)-\left( 1-\frac{d}{r} \right) \right]=\frac{2Qd \, \hat r}{4\pi\epsilon_0 r^3} $$

Electric field perpendicular to dipole

Now this is the situation:

• In $\hat \theta$ direction:

$$ \vec E=2\frac{Q\,\hat \theta}{4\pi\epsilon_0 r^2}\,\sin\alpha $$

• For $r \gg d$ so $\alpha$ is small

$$ \sin\alpha =\frac{d/2}{r}\Rightarrow\vec E=\frac{Qd\,\hat \theta}{4\pi\epsilon_0 r^3} $$

General potential from a dipole

🌟 For charges

$$ \phi=\phi_++\phi_-=\frac{Q}{4\pi\epsilon_0 r_1}-\frac{Q}{4\pi\epsilon_0 r_2} $$

where $r_1=r-\frac{d}{2}\cos\theta$ and $r_2=r+\frac{d}{2}\cos\theta$

• Approximate:

$$ \begin{aligned}\frac{1}{r_1}=\frac{1}{r-\frac{d}{2}\cos\theta}=\frac{1}{r\,\left(1-\frac{d\cos\theta}{2r}\right)}\approx\frac{1+\frac{d\cos\theta}{2r}}{r}\end{aligned} $$

thus we get

$$ \phi\approx\frac{Qd\cos\theta}{4\pi\epsilon_0 r^2} $$

⚡ For electric field

$$ \vec E=-\vec\nabla \phi=-\left( \frac{\partial \phi}{\partial r}\, \hat r+\frac{1}{r}\frac{\partial \phi}{\partial \theta}\,\hat \theta \right) \quad ; \quad \phi=\frac{Qd\cos\theta}{4\pi\epsilon_0 r^2} \\ \begin{aligned} \vec E &= -\frac{Qd}{4\pi\epsilon_0}\left[\frac{\partial}{\partial r} \left(\frac{\cos\theta}{r^2}\right)\, \hat r + \frac{1}{r}\frac{\partial}{\partial \theta}\left(\frac{\cos\theta}{r^2}\right)\,\hat\theta\right] \\ \vec E&=-\frac{Qd}{4\pi\epsilon_0}\left [ \left(\frac{-2\cos\theta}{r^3}\right)\,\hat r+\frac{1}{r}\left( \frac{-\sin\theta}{r^2} \right) \right]\\ \vec E&=\frac{Qd}{4\pi\epsilon_0r^3}(2\cos\theta\,\hat r+\sin\theta\,\hat\theta) \end{aligned} $$

Energy and torque

A dipole in a uniform E-field

• Force $\vec F_+=q\vec E$ and $\vec F_-=-q\vec E$ on two charges, no net force
• Torque is $\vec \tau=\vec p \times \vec E$