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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 2: Electrostatics II: Field Lines and Continuous Charges</b><br>
Thu 22 Jan 2026 11:00
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[[Year 1 Sem 2/LC Electromagnetism I/Lecture 01|← Previous lecture]] · [[Year 1 Sem 2/LC Electromagnetism I|Module]] · [[Year 1 Sem 2/LC Electromagnetism I/Lecture 03|Next lecture →]]

== Field Lines ==

To visualise a field, we draw field lines. These are drawn so that the tangent of these lines at any point in spaces gives the direction of <nowiki>$\underline{E}$</nowiki> at that point.

The lines point radially outwards from a positive charge, and radially towards a negative charge. The density of these lines is proportional to the magnitude of <nowiki>$\underline{E}$</nowiki>, but the absolute number of lines doesn’t convey much.

Field lines must start at a positive charge and end on a negative charge. They are continuous and can never cross each other, they only start/stop at charges or proceed to infinity.

== Continuous Charge Distributions ==

What do we do if we have a charge which isn’t a point particle?

<ul>
<li>

We break the object up into a distribution of infinitesimal point charges.

</li>
<li>

We treat each piece as a point charge and add up the separate <nowiki>$\underline{E}$</nowiki> fields across an integral.

</li>
<li>

This gets messy for anything non-trivial, resulting in a complex 3D integral often requiring computational techniques.

</li>
</ul>

Charges can be distributed across an object in a number of ways:

<ul>
<li>

Charge per unit length across a line, given by <nowiki>$\lambda$</nowiki>: <div class="pn-import-equation"><nowiki>$$q = \int \lambda dl$$</nowiki></div>

</li>
<li>

Charge per unit area across a surface area, given by <nowiki>$\sigma$</nowiki>: <div class="pn-import-equation"><nowiki>$$q = \int \sigma dA$$</nowiki></div>

</li>
<li>

Charge per unit volume, given by <nowiki>$\rho$</nowiki>: <div class="pn-import-equation"><nowiki>$$q = \int \rho dV$$</nowiki></div>

</li>
</ul>

== Example: E-Field from a Line of Charge ==

Consider an E-field from a line of length <nowiki>$2L$</nowiki>, measured at a point along the midpoint’s perpendicular. This point sits <nowiki>$x$</nowiki> length units along the perpendicular.

[[File:EM-Y1S2-lec02-01.png|thumb|center|600px]]

We consider some element of charge <nowiki>$\delta Q = \lambda \delta y$</nowiki> along the line at y. There are two components to the E-field, one pointing in the positive x direction and one in the negative y direction.

When we consider all of the <nowiki>$\delta Q$</nowiki>s along the length of the charged rod, each point on the positive y-axis along the length will have a corresponding point on the negative y-axis. These two points will have equal and opposite magnitude, so <nowiki>$E_y = 0$</nowiki> by symmetry. There is therefore no resultant <nowiki>$y$</nowiki> component and the final field will point along <nowiki>$\hat{x}$</nowiki>.

For some infinitesimal <nowiki>$\delta q$</nowiki> generically along the line of charge: <div class="pn-import-equation"><nowiki>$$\delta E_x = \frac{\delta q}{4 \pi \epsilon_0 r^2} \cos \alpha \tag{1}$$</nowiki></div>

And for each <nowiki>$\delta Q$</nowiki>: <div class="pn-import-equation"><nowiki>$$\delta Q = \lambda \delta y \tag{2}$$</nowiki></div>

And from trig: <div class="pn-import-equation"><nowiki>$$r^2 = y^2 + x^2 \tag{3}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\cos \alpha = \frac{x}{r} \tag{4}$$</nowiki></div>

Substituting 2, 3, 4 into 1:

<div class="pn-import-equation"><nowiki>$$\delta E_x = \frac{\lambda \delta y}{4 \pi \epsilon_0} \frac{x}{\left[y^2 + x^2\right]^{3/2}}$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$E_x = \frac{\lambda x}{4 \pi \epsilon 0} \int_{-L}^{L} \frac{1}{\left[y^2 + x^2\right]^{3/2}} \, dy$$</nowiki></div>

To solve this, we would use the standard integral given in the question. Alternatively, if a standard integral isn’t given in the question we could express in the integral in terms of <nowiki>$\alpha$</nowiki> and integrate from the minimum to maximum alpha.

=== Example Part B: The Easier Route ===

As before: <div class="pn-import-equation"><nowiki>$$\delta E_x = \frac{\lambda \delta y}{4 \pi \epsilon_0 r^2} \cos \alpha \tag{1}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\cos \alpha = \frac{x}{r} \implies r = \frac{x}{\cos \alpha} \tag{2}$$</nowiki></div>

And: <div class="pn-import-equation"><nowiki>$$\tan \alpha = \frac{y}{x} \implies y = x \tan \alpha \tag{3}$$</nowiki></div>

As: <div class="pn-import-equation"><nowiki>$$\frac{d}{d \alpha} \tan \alpha = \frac{1}{\cos^2 \alpha}$$</nowiki></div> We have: <div class="pn-import-equation"><nowiki>$$\delta y = \frac{x \delta \alpha}{\cos^2 \alpha} \tag{4}$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$\delta E_x = \frac{\lambda}{4 \pi \epsilon_0} \underbrace{\frac{x \delta x}{\cos^2 \alpha}}_\text{$\delta y$} \underbrace{\frac{\cos^2 \alpha}{x^2}}_\text{$1/r^2$} \cos \alpha$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{\lambda}{4 \pi \epsilon_0 x} \cos \alpha$$</nowiki></div>

Which is much easier to integrate from the minimum to maximum values of <nowiki>$\alpha$</nowiki>.

<div class="pn-import-equation"><nowiki>$$E_x = \frac{\lambda}{4 \pi \epsilon_0 x} \int_{\alpha_\text{min}}^{\alpha_\text{max}} \cos \alpha \, d \alpha$$</nowiki></div>

And from the geometry of the problem: <div class="pn-import-equation"><nowiki>$$\alpha_\text{max} = \arctan \left(\frac{L}{x}\right) \qquad \alpha_\text{min} = - \arctan \left(\frac{L}{x}\right)$$</nowiki></div>

Which is easy to integrate.

If the problem was setup such that the rod was infinitely long, then: <div class="pn-import-equation"><nowiki>$$E_x = \frac{\lambda}{4 \pi \epsilon_0 x} \int_{-\pi / 2}^{\pi / 2} \cos \alpha \, d \alpha$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$E_x = \frac{\lambda}{2 \pi \epsilon_0 x}$$</nowiki></div>

This is approximately true for the case where <nowiki>$x \ll L$</nowiki>.

== Example: Ring of Charge ==

Consider a ring of charge with radius <nowiki>$a$</nowiki>. We step out some perpendicular constant distance <nowiki>$x$</nowiki> from the centre of the ring to point <nowiki>$P$</nowiki>. What is the resultant field at <nowiki>$P$</nowiki>?

[[File:EM-Y1S2-lec02-02.png|thumb|center|600px]]

The ring has total charge <nowiki>$Q$</nowiki>, distributed with charge per unit length <nowiki>$\lambda$</nowiki>.

Again, we can use symmetry. For every point on the ring, there will be a point one diameter away creating an equal and opposite <nowiki>$\delta E$</nowiki>. Hence, the <nowiki>$y$</nowiki> and <nowiki>$z$</nowiki> components cancel leaving only <nowiki>$E_x$</nowiki>, pointing in the <nowiki>$\hat{x}$</nowiki> direction.

<div class="pn-import-equation"><nowiki>$$E_y = E_z = 0 \text{ by symmetry}$$</nowiki></div>

We have: <div class="pn-import-equation"><nowiki>$$\delta E_x = \frac{\delta q}{4 \pi \epsilon_0 r^2} \cos \theta$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\cos \theta = \frac{x}{y}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$r = \sqrt{a^2 + x^2}$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$\delta E_x = \frac{\delta q}{4 \pi \epsilon_0} \frac{x}{\left[a^2 + x^2\right]^{3/2}}$$</nowiki></div>

Here, a and x are constants, so the only variable we can integrate across is <nowiki>$\delta q$</nowiki>: <div class="pn-import-equation"><nowiki>$$E_x = \frac{x}{4 \pi \epsilon_0} \frac{1}{\left[a^2 + x^2\right]^{3/2}} \int dq$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{x Q}{4 \pi \epsilon_0} \frac{1}{\left[a^2 + x^2\right]^{3/2}}$$</nowiki></div>

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