Electric and Magnetic Fields

Point Charge

A point charge is a point particle with nonzero electric charge and infinitesimal volume.

Coulombs Law

Also called Coulomb's inverse-square law states the magnitude of the attraction or repulsion electrostatic force ($\bm{F}$) between two charges is directly proportional to the magnitude of charges ($q$) and inversely proportional to the square of the magnitude of the separation displacement ($\bm{r}$):

$$\begin{aligned} |\bm{F}| \propto |q_1||q_2|, \quad |\bm{F}| \propto |\bm{r}|^2\\ |\bm{F}| = k \frac{|q_1||q_2|}{|\bm{r}|^2} \end{aligned}$$

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Coulomb's Law can also be expressed in vector form, where the force exerted by $q_1$ on $q_2$ is:

$$\begin{aligned} \bm{F_{12}} = k \frac{q_1 q_2}{|r|^2} \hat{r_{12}} = -\bm{F_{21}} \end{aligned}$$

Where the constant $k$ is named Coulomb's constant and is equal to:

$$\begin{aligned} k = \frac{1}{4\pi \epsilon_0} \end{aligned}$$

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The derivation of Coulomb's constant can be possible with Maxwell's equations. Gauss's Law states that the electric flux ($\Phi$) through any closed surface is equal to net electric charge enclosed within the surface multiply by the inverse of permittivity of free space ($\epsilon_0$):

$$\begin{aligned} \Phi = \oiint\limits_{S} E \cdot dA = \frac{Q}{\epsilon_0} \end{aligned}$$

For a point charge, symmetry implies a radial electric field, and hence a sphere enclosed surface ($4\pi r^2$), therefore, simplifying $\phi$ to:

$$\begin{aligned} \Phi = 4\pi r^2 = \frac{Q}{\epsilon_0}\\ E = \left(\frac{1}{4\pi\epsilon_0}\right)\left(\frac{Q}{r^2}\right)\\ F = q_2 E = \frac{1}{4\pi \epsilon_0} \frac{|q_1||q_2|}{r^2} \end{aligned}$$

Coulomb's constant can now be identify as:

$$\begin{aligned} k = \frac{1}{4\pi \epsilon_0} \end{aligned}$$

Electric Field

The electric field $\bm{E}$ with vector quantity of electrostatic force $\bm{F}$ exerted per unit of magnitude of charge, evaluated as:

$$\begin{aligned} \bm{E} = \lim_{q \rightarrow 0} \frac{F}{q} \end{aligned}$$

Electric Potential Energy

Electric potential energy ($U$ or $E_p$ or $W$)is the minimum work required to translate charges from infinite separation displacement to a position $\bm{r}$. Since the electrostatic force $\bm{F_e}$ varies due to the displacement, the line integral of the opposing force $\bm{F_{ext}} = \bm{F_e}$ from infinity $(\infty)$ to the potion $\bm{r}$ gives the work required to translate the charges:

$$\begin{aligned} U &= E_p = W = \int_\infty^{\bm{r}} F_{ext} \cdot d\bm{r} = -\int_\infty^{\bm{r}} F_e \cdot \bm{dr}\\ U &= -\int_\infty^{\bm{r}} \left(k\frac{q_1 q_2}{|\bm{r}|^2}\bm{\hat{r}}\right) \cdot \bm{dr} \end{aligned}$$

Separating the magnitude and direction of $\bm{dr}$ gives $d|\bm{r}|\bm{\hat{r}}$, and hence changing the bounds ($\bm{r} \rightarrow |\bm{r}|$):

$$\begin{aligned} U &= -\int_\infty^{|\bm{r}|} \left(k\frac{q_1 q_2}{|\bm{r}|^2}\bm{\hat{r}}\right) \cdot d|\bm{r}|\bm{\hat{r}}\\ U &= -\frac{q_1 q_2}{4\pi\epsilon_0} \int_\infty^{|\bm{r}|} \frac{1}{|\bm{r}|^2} d|\bm{r}|\\ U &= -\frac{q_1 q_2}{4\pi\epsilon_0} \left[-\frac{1}{|\bm{r}|}\right]_\infty^{|\bm{r}|}\\ U &= \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{|\bm{r}|} \end{aligned}$$

This is written in the formula booklet with $r = |\bm{r}|$ and Coulomb's constant $k$:

$$\begin{aligned} U = k\frac{q_1 q_2}{r} \end{aligned}$$

This implies the electric potential energy is a scalar quantity approaching $0$ at infinite displacement $r \rightarrow \infty$.

Electric Potential

The electric potential ($V_e$) is the electric potential energy of a point charge ($Q$ or $q_1$) at displacement $\bm{r}$ per unit charge:

$$\begin{aligned} V_e &= \frac{U}{q_2} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{|\bm{r}|q_2} = \frac{1}{4\pi \epsilon_0}\frac{Q}{|\bm{r}|}\\ V_e &= \frac{kQ}{r}, \quad r = |\bm{r}| \end{aligned}$$

Electric Field Strength

The electric field strength ($\bm{E}$) is the force ($\bm{F_{12}}$) per unit charge of the affected point charge ($q_2$) at a displacement ($\bm{r_{12}}$):

$$\begin{aligned} \bm{E} = \frac{\bm{F_{12}}}{q_2} = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{|\bm{r}|^2}\bm{\hat{r_{12}}} \end{aligned}$$

This is written in the formula booklet with $d = |\bm{r}|$:

$$\begin{aligned} E = k\frac{F}{d} \end{aligned}$$

The electric field strength is also the negative gradient of the electric potential ($V_e$):

$$\begin{aligned} E = -\nabla V_e \end{aligned}$$

Since spherical symmetry implies the gradient of electric potential are equal for equal magnitude of displacement $\bm{r}$:

$$\nabla V_e = \frac{dV_e}{d|\bm{r}|}\bm{\hat{r}}\\ \frac{dV_e}{d|\bm{r}|} = \frac{d}{d|\bm{r}|}\left(\frac{kQ}{|\bm{r}|}\right) = -k\frac{Q}{|\bm{r}|^2}\\ E = k \frac{Q}{|\bm{r}|^2}\bm{\hat{r}}$$

When expressing the electric field strength with the average electric potential change, and expressing $r = |\bm{r}|$, the magnitude of $E$ is:

$$\begin{aligned} E = -\frac{\Delta V_e}{\Delta r} \end{aligned}$$

Graphical Presentation of Electric Field

Electric Field Lines

Electric field lines present the direction of electric field strength.

Equipotential Surfaces

Equipotential surfaces are surfaces with equivalent electric potential.

Millikan's Experiment

The Millikan's experiment was conducted in 1909 to determine the value of elementary charge. The experiment was passing ionized oil drops with charge $q$ within a region between two charged metal plates with a electric potential $V_e$, and displacement $\bm{d}$. The IB include a simplified calculation for elementary charge by ignoring buoyance force, where the electrostatic force ($\bm{F_e}$) equal to the opposite of gravitational force ($\bm{F_g}$):

$$\begin{aligned} \bm{F_g} &= -\bm{F_e}\\ \bm{F_g} &= m\bm{g}\\ -\bm{F_e} &= -qE = -q\left(-\frac{V_e}{\bm{d}}\right)\\ m\bm{g} &= q\frac{V_e}{\bm{d}}\\ q &= \frac{m\bm{g}\bm{d}}{V_e} \end{aligned}$$