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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 11: Electrostatics XI: Revision</b><br>
Mon 23 Feb 2026 11:00
</div>

[[Year 1 Sem 2/LC Electromagnetism I/Lecture 10|← Previous lecture]] · [[Year 1 Sem 2/LC Electromagnetism I|Module]] · [[Year 1 Sem 2/LC Electromagnetism I/Lecture 12|Next lecture →]]

== Example I ==

Two point charges <nowiki>$+Q$</nowiki> and <nowiki>$-Q$</nowiki> are at <nowiki>$(x, y)$</nowiki> coordinates <nowiki>$(-2a, 0)$</nowiki> and <nowiki>$2a, 0$</nowiki> respectively.

What is the potential at <nowiki>$(2a, 3a)$</nowiki>? We work out the contributions in potential from each charge and sum them.

For the positive charge: <div class="pn-import-equation"><nowiki>$$V = \frac{Q}{4 \pi \epsilon_0 \sqrt{(3a)^2 + (4a)^2}} = \frac{Q}{4 \pi \epsilon_0 5a}$$</nowiki></div>

And the negative charge: <div class="pn-import-equation"><nowiki>$$V = -\frac{Q}{4 \pi \epsilon_0 3a}$$</nowiki></div>

Hence the total: <div class="pn-import-equation"><nowiki>$$V_T = \frac{Q}{4 \pi \epsilon_0 a} \left[\frac{1}{5} - \frac{1}{3}\right]$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= - \frac{Q}{4 \pi \epsilon_0 a} \frac{2}{15}$$</nowiki></div>

What is the work done to move the charge from <nowiki>$(-2a, 3a)$</nowiki> to <nowiki>$(-2a, -3a)$</nowiki>?

<div class="pn-import-equation"><nowiki>$$W = q \Delta V$$</nowiki></div>

Where the initial potential is: <div class="pn-import-equation"><nowiki>$$V(-2a, 3a) = \frac{Q}{4 \pi \epsilon_0 (3a)} - \frac{Q}{4 \pi \epsilon_0 (5a)}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{Q}{4 \pi \epsilon_0 a} \frac{2}{15}$$</nowiki></div>

And the final potential is: <div class="pn-import-equation"><nowiki>$$V(-2a, -3a) = \frac{Q}{4 \pi \epsilon_0 (5a)} - \frac{Q}{4 \pi \epsilon_0 (3a)}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= - \frac{Q}{4 \pi \epsilon_0 a} \frac{2}{15}$$</nowiki></div>

Hence the work is given by: <div class="pn-import-equation"><nowiki>$$W = q \Delta V = q \left( V(-2a, 3a) - V(-2a, -3a) \right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= q \left( \frac{Q}{4 \pi \epsilon_0 a} \frac{2}{15} - \left(- \frac{Q}{4 \pi \epsilon_0 a} \frac{2}{15} \right) \right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{4qQ}{15 \cdot 4 \pi \epsilon_0 a}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{qQ}{15 \pi \epsilon_0 a}$$</nowiki></div>

Since we’re going from a higher potential to a lower potential, the work done is positive, as expected. This corresponds to work being done <i>by</i> the electric field.

== Example II ==

An infinitely long hollow cylinder has internal radius <nowiki>$R_1$</nowiki> and external radius <nowiki>$R_2$</nowiki>. For <nowiki>$R_1 < r < R_2$</nowiki>, the cylinder has charge density <nowiki>$\rho(r) =\rho_0 r$</nowiki>, and for <nowiki>$r < R_1$</nowiki> and <nowiki>$r > R_2$</nowiki>, <nowiki>$\rho(r) = 0$</nowiki>.

<ol>
<li>

Show that the total charge <nowiki>$Q$</nowiki> per unit length of the cylinder is: <div class="pn-import-equation"><nowiki>$$Q = \frac{2}{3} \pi \rho_0 \left(R_2^3 - R_1^3\right)$$</nowiki></div>

</li>
<li>

Calculate the electric field at radius <nowiki>$r$</nowiki> for:

<ol>
<li>

<nowiki>$r > R_2$</nowiki>

</li>
<li>

<nowiki>$R_1 < R < R_2$</nowiki>

</li>
</ol>

Expressing in terms of <nowiki>$r$</nowiki> and <nowiki>$\rho_0$</nowiki>.

Sketch a graph to show how the field varies with <nowiki>$r$</nowiki>.

</li>
<li>

Use the previous result to derive an expression for <nowiki>$V(r)$</nowiki> for <nowiki>$r < R_2$</nowiki>

</li>
</ol>

<b>Answer 1</b>

Consider a cylindrical shell at radius <nowiki>$r$</nowiki> and thickness <nowiki>$\delta r$</nowiki>. Because the shell has infinitesimal thickness, we assume that charge density is constant throughout. <div class="pn-import-equation"><nowiki>$$\delta Q = \rho_0 r \times 2 \pi r dr \times L$$</nowiki></div> And at unit length <nowiki>$L = 1$</nowiki> <div class="pn-import-equation"><nowiki>$$\delta Q = \rho_0 r \times 2 \pi r dr$$</nowiki></div>

Hence we can integrate for the total charge: <div class="pn-import-equation"><nowiki>$$Q = \int_{R_1}^{R_2} \rho_0 2 \pi r^2 \, dr$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= 2 \pi \rho_0 \left(\frac{R_2^3}{3} - \frac{R_1^3}{3}\right)$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$Q = \frac{2}{3} \pi \rho_0 \left(R_2^3 - R_1^3\right)$$</nowiki></div>

<b>Answer 2</b>

We use a cylindrical Gaussian surface of radius <nowiki>$r$</nowiki>. Ignoring edge effects, the electric field is constant on the surface and directed radially outwards. Hence <nowiki>$\vec{E}$</nowiki> and <nowiki>$d \vec{S}$</nowiki> are parallel, and we can write: <div class="pn-import-equation"><nowiki>$$\int_S \vec{E} \cdot d \vec{S} = \frac{Q_\text{enc}}{\epsilon_0}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$ES = \frac{Q_\text{enc}}{\epsilon_0}$$</nowiki></div>

Initially taking it outside the cylinder, <nowiki>$r > R_2$</nowiki>: <div class="pn-import-equation"><nowiki>$$\text{LHS} = 2 \pi r L E$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\text{RHS} =\frac{1}{\epsilon_0} \frac{2}{3} \pi \rho_0 \left(R_2^3 - R_1^3\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies E = \frac{\rho_0}{3 \epsilon_0} \frac{R_2^3 - R_1^3}{r} \text{ radially}$$</nowiki></div>

For <nowiki>$R_1 < r < R_2$</nowiki>: <div class="pn-import-equation"><nowiki>$$\text{LHS} = 2 \pi r L E$$</nowiki></div>

The RHS contains a smaller volume of charge, so we integrate up to <nowiki>$r$</nowiki> instead of <nowiki>$R_2$</nowiki>: <div class="pn-import-equation"><nowiki>$$\text{RHS} = \frac{1}{\epsilon_0} \int_{R_1}^r \rho_0 2 \pi r^2 dr$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{2 \pi \rho_0 L}{3 \epsilon_0} \left(r^3 - R_1^3\right)$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$2 \pi r L E = \frac{2 \pi \rho_0 L}{3 \epsilon_0} \left(r^3 - R_1^3\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$E = \frac{\rho_0}{3 \epsilon_0} \frac{r^3 - R_1^3}{r} = \frac{\rho_0}{3 \epsilon_0} \left(r^2 - \frac{R_1^3}{r}\right) \text{ radially}$$</nowiki></div>

To sketch, we get:

<ul>
<li>

No enclosed charge, hence no field for <nowiki>$r < R_1$</nowiki>.

</li>
<li>

Cubic growth from <nowiki>$R_1 < r < R_2$</nowiki>.

</li>
<li>

<nowiki>$1/r$</nowiki> decat <nowiki>$r > R_2$</nowiki>.

</li>
</ul>

<b>Answer 3</b>

For <nowiki>$r > R_2$</nowiki>, we have determined: <div class="pn-import-equation"><nowiki>$$E = \frac{\rho_0}{3 \epsilon_0} \frac{R_2^3 - R_1^3}{r}$$</nowiki></div>

So: <div class="pn-import-equation"><nowiki>$$V = - \frac{\rho_0}{3 \epsilon_0} (R_2^3 - R_1^3) \int \frac{1}{r} dr$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= - \frac{\rho_0}{3 \epsilon_0} (R^3_2 - R^3_1) \ln r + c$$</nowiki></div>

As this doesn’t tend to zero as <nowiki>$r \to \infty$</nowiki>, we cannot determine the constant <nowiki>$c$</nowiki>. We would need a reference point at some finite <nowiki>$r$</nowiki> to determine <nowiki>$c$</nowiki>, but we don’t have that information, so we leave it as is.

== Example III ==

A wire is bent into a semicircle of radius <nowiki>$R$</nowiki>. It has uniform charge density <nowiki>$+ \lambda$</nowiki> per unit length. What is the magnitude and direction of the electric field at the centre of the semicircle?

Because it’s not symmetric, we can’t use Gauss’s law and need to use the integration method. We take a small arc length of the wire <nowiki>$\delta \theta$</nowiki> subtending a charge <nowiki>$\delta q$</nowiki>. This creates a field at <nowiki>$\delta \theta$</nowiki> from the horizontal. For each horizontal component in one direction, there is an equal and opposite component in the opposite direction, so the horizontal components cancel out. Hence by symmetry the resultant direction is the negative vertical and we can only consider magnitude.

<div class="pn-import-equation"><nowiki>$$\delta E_y = \frac{\delta Q}{4 \pi \epsilon_0 R^2} = \frac{\lambda \delta L}{4 \pi \epsilon_0 R^2}$$</nowiki></div>

And as the subtended arc length is <nowiki>$\delta L = R \delta \theta$</nowiki>, <div class="pn-import-equation"><nowiki>$$\delta E = \frac{\lambda R \delta \theta}{4 \pi \epsilon_0 R^2}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\delta E = \frac{\lambda \delta \theta}{4 \pi \epsilon_0 R}$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$\delta E_y = \delta E \sin \theta$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\delta E_y = \frac{lambda}{4 \pi \epsilon_0 R} \sin \theta \delta \theta$$</nowiki></div>

And integrating across the semicircle: <div class="pn-import-equation"><nowiki>$$E_y = \int_{0}^{\pi} \frac{\lambda}{4 \pi \epsilon_0 R} \sin \theta \, d \theta$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{\lambda}{4 \pi \epsilon_0 R} \int_0^\pi \sin \theta \, d \theta$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{\lambda}{4 \pi \epsilon_0 R} \left[ -\cos \theta \right]_0^\pi$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{\lambda}{4 \pi \epsilon_0 R} \left( -\cos \pi + \cos 0 \right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{\lambda}{4 \pi \epsilon_0 R} \left( 1 + 1 \right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \frac{\lambda}{2 \pi \epsilon_0 R}$$</nowiki></div>

And the direction is vertically downward (negative y-direction).

<div class="pn-import-equation"><nowiki>$$\vec{E} = \frac{- \lambda}{2 \pi \epsilon_0 R} \hat{y}$$</nowiki></div>

== Example IV ==

A ring shaped conductor of radius <nowiki>$a$</nowiki> carries a total charge <nowiki>$Q$</nowiki> uniformly distributed around it. The ring is in the <nowiki>$(y, z)$</nowiki> plane, centred on the origin.

<ol>
<li>

Find the potential at a point <nowiki>$P$</nowiki> on the ring axis at a distance <nowiki>$x$</nowiki> from the centre of the ring.

</li>
<li>

Hence show that the only non-zero component of the E-field at P is along the ring axis and is given by: <div class="pn-import-equation"><nowiki>$$E_x = \frac{Qx}{4 \pi \epsilon_0 (a^2 + x^2)^{3/2}}$$</nowiki></div>

</li>
<li>

If an electron is placed at the centre of the ring and is displaced a small distance <nowiki>$x$</nowiki> where <nowiki>$x \ll a$</nowiki>. Show that the electron oscillates with a frequency <nowiki>$f$</nowiki>.

</li>
</ol>

<b>Answer 1</b>

Let <nowiki>$r = \sqrt{a^2 + x^2}$</nowiki>.

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

<b>Answer 2</b> Using <nowiki>$E = - \nabla V$</nowiki>: <div class="pn-import-equation"><nowiki>$$\vec{E} = - \nabla V = - \frac{\partial V}{\partial x} \hat{x} - \frac{\partial V}{\partial y} \hat{y} - \frac{\partial V}{\partial z} \hat{z}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\frac{\partial V}{\partial y} = \frac{\partial V}{\partial z} = 0$$</nowiki></div> Hence there is no field in the <nowiki>$y$</nowiki> and <nowiki>$z$</nowiki> directions, and the only non-zero component is along the <nowiki>$x$</nowiki>-axis. <nowiki>$E_y = E_z = 0$</nowiki>.

So: <div class="pn-import-equation"><nowiki>$$\vec{E} = - \frac{\partial}{\partial x} \left(\frac{Q}{4 \pi \epsilon_0 \sqrt{a^2 + x^2}}\right) \hat{x}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\vec{E} = \frac{Q}{4 \pi \epsilon_0} \left(- \frac{1}{2} \times 2x (a^2 + x^2)^{-3/2}\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\vec{E} = \frac{Qx}{4 \pi \epsilon_0 (a^2 + x^2)^{3/2}} \hat{x}$$</nowiki></div>

<b>Answer 3</b>

We can rewrite: <div class="pn-import-equation"><nowiki>$$E = \frac{Q}{4 \pi \epsilon_0} \frac{x}{a^3 \left(1 + \frac{x^2}{a^2} \right)^{3/2}}$$</nowiki></div>

For <nowiki>$x \ll a$</nowiki>, we can approximate <sup>[[#EM-11-note-1|1]]</sup>: <div class="pn-import-equation"><nowiki>$$E \approx \frac{Qx}{4 \pi \epsilon_0 a^3}$$</nowiki></div>

And the force is: <div class="pn-import-equation"><nowiki>$$\vec{F} = -e \vec{E} = \frac{-e Qx}{4 \pi \epsilon_0 a^3} \hat{x}$$</nowiki></div>

Using <nowiki>$F = ma$</nowiki>: <div class="pn-import-equation"><nowiki>$$m \frac{d^2 x}{dt^2} = \frac{-e Qx}{4 \pi \epsilon_0 a^3}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\frac{d^2 x}{dt^2} + \frac{e Q}{4 \pi \epsilon_0 m a^3} x = 0$$</nowiki></div> This is the equation of a simple harmonic oscillator: <div class="pn-import-equation"><nowiki>$$x(t) = x_0 \cos \omega t$$</nowiki></div> Where: <div class="pn-import-equation"><nowiki>$$\omega^2 = \frac{e Q}{4 \pi \epsilon_0 m a^3}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\omega = \sqrt{\frac{e Q}{4 \pi \epsilon_0 m a^3}}$$</nowiki></div>

And finally <nowiki>$\omega = 2 \pi f$</nowiki>: <div class="pn-import-equation"><nowiki>$$f = \frac{1}{2 \pi} \sqrt{\frac{e Q}{4 \pi \epsilon_0 m a^3}}$$</nowiki></div>

== Footnotes ==

<div id="EM-11-note-1"><b>1.</b>

Note that if the <nowiki>$x$</nowiki> term wasn’t present in the numerator, we’d have to do a first-order binomial expansion, but as <nowiki>$x$</nowiki> is present, we can just approximate the denominator to <nowiki>$a^3$</nowiki>.

</div>

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Year 1 Sem 2/LC Electromagnetism I/Lecture 11 - Revision history

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