Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 7: Electrostatics VII: Electric Dipoles</b><br>
Mon 10 Feb 2026 12:00
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== Using Potential to Determine E ==

Consider two charges <nowiki>$Q$</nowiki> with distance <nowiki>$L$</nowiki> between them. The potential at some generic <nowiki>$x$</nowiki> is:

<div class="pn-import-equation"><nowiki>$$V(x) = \frac{Q}{4 \pi \epsilon_0 x} + \frac{Q}{4 \pi \epsilon_0 (L -x)}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\pmb{E} = - \nabla V = - \frac{dV}{dx} \hat{\pmb{x}}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \left(\frac{Q}{4 \pi \epsilon_0 x^2} - \frac{Q}{4 \pi \epsilon_0 (L - x)^2}\right) \hat{\pmb{x}}$$</nowiki></div>

At the middle point <nowiki>$x = L/2$</nowiki>: <div class="pn-import-equation"><nowiki>$$V(x = L/2) = \frac{Q}{4 \pi \epsilon_0 L/2} + \frac{Q}{4 \pi \epsilon_0 (L - L/2)}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\pmb{E}(x = L/2) = \frac{Q}{4 \pi \epsilon_0 L^2/4} - \frac{Q}{4 \pi \epsilon_0 (L^2 / 4)} = 0$$</nowiki></div>

== Potential from a Ring ==

[[File:EM-Y1S2-lec07-01.png|thumb|center|600px]]

A ring-shaped conductor of radius <nowiki>$a$</nowiki> carries a total charge <nowiki>$Q$</nowiki> uniformly distributed around it. Find the potential, and hence the electric field, at point <nowiki>$P$</nowiki>.

Consider some small element of charge <nowiki>$\delta Q$</nowiki>, at distance <nowiki>$r = \sqrt{a^2 + x^2}$</nowiki> from <nowiki>$P$</nowiki>.

<div class="pn-import-equation"><nowiki>$$\delta V = \frac{\delta Q}{4 \pi \epsilon_0 r} = \frac{\delta Q}{4 \pi \epsilon_0 (a^2 + x^2)^{1/2}}$$</nowiki></div>

We can then integrate, noting that <nowiki>$a, x$</nowiki> are both constants: <div class="pn-import-equation"><nowiki>$$V = \frac{1}{4 \pi \epsilon_0 (a^2 + x^2)^{1/2}} \oint dQ$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{Q}{4 \pi \epsilon_0 (a^2 + x^2)^{1/2}}$$</nowiki></div>

And then to find <nowiki>$\pmb{E}$</nowiki>, noting that <nowiki>$\pmb{E} = - \nabla V$</nowiki> (and that the derivatives in the z, y directions are zero): <div class="pn-import-equation"><nowiki>$$\pmb{E} = - \frac{\partial V}{\partial x} \pmb{\hat{x}} = - \frac{Q}{4 \pi \epsilon_0} \left(- \frac{1}{2} 2x (a^2 + x^2)^{-3/2}\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{Qx}{4 \pi \epsilon_0 \left(a^2 + x^2\right)^{-3/2}} \hat{\pmb{x}}$$</nowiki></div>

== Electric Dipoles ==

Consider two charges <nowiki>$+q$</nowiki> and <nowiki>$-q$</nowiki>, separated by a distance <nowiki>$a$</nowiki>. This forms an electric dipole. This has an associated quantity given the electric dipole moment: <div class="pn-import-equation"><nowiki>$$\pmb{p} = q \pmb{a}$$</nowiki></div> We define this vector to point against the -field, i.e. pointing from the negative towards the positive charge.

Many molecules naturally form dipoles. Because water is not vertically symmetric, it causes a difference between the effective positive charge from the two hydrogens and the net negative charge from the oxygen.

This makes it an excellent solvent and is used in heating food via microwaves. Because the distance is very small, it gives a very small value of <nowiki>$p = 6.1 \times 10^{-30}$</nowiki>Cm per molecule.

[[File:EM-Y1S2-lec07-02.png|thumb|center|400px]]

For a dipole, the potential is given by: <div class="pn-import-equation"><nowiki>$$V_p \approx \frac{p \cos \theta}{4 \pi \epsilon_0 r^2}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\pmb{E} \approx \frac{2 p \cos \theta}{4 \pi \epsilon_0 r^3} \hat{\pmb{r}} + \frac{p \sin \theta}{4 \pi \epsilon_0 r^3} \hat{\pmb{\theta}}$$</nowiki></div> For <nowiki>$r \gg a$</nowiki> and where <nowiki>$\theta$</nowiki> is defined as the angle between <nowiki>$\pmb{r}$</nowiki> and the positive charge.s

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Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc