PHYS 152
PHYS 152 | Electricity and Magnetism
Chapter 23 | Electric Fields
Lecture 2022-06-29 ✅
The Coulomb ($\rm C$) is the unit of electric charge. Coulomb's constant, $k_e$
Coulomb's Law: Given two particles with charges $q_1$ and $q_2$ respectively, separated by a distance of $r$, the electric force $\vec{F_e}$ experienced by each particle is defined by:
$$
\vec{F_e} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_1 q_2}{r^2} \hat{r}_{12}
$$
Where $\varepsilon_0$ is the electric constant, (formerly known as the permittivity of free space or vacuum permittivity), a constant defined as:
$$
\varepsilon_0 \approx 8.85 * 10^{-12} ~ \rm \frac{C^2}{N \cdot m^2}
$$
It is common to represent this equation using Coulomb's constant, $k_e$. though, since the 2019 redefinition, the Coulomb constant is no longer exactly defined. Nevertheless, the substitution is as follows:
$$
k_e = \frac{1}{4\pi \varepsilon_0} \approx 8.99 * 10^{9} ~ \rm \frac{N \cdot m^2}{C^2}
$$
It's worth noting that the units of $k_e$, when multiplied against the rest of the formula causes many units to be cancelled out, resulting in electric force being defined in terms of Newtons ($N$)
- $\rm m^2$ cancelled out by the units in $r^2$
- $\rm C^{2}$ cancelled out by the units in $q_1$ and $q_2$.
As a result, the electric force, as it is defined in this formula, is ultimately in terms of $\rm N$ (Newtons, the unit of force).
Lecture 2022-06-30 ✅
A charge $q$ produces an electric field $\vec{E}$ around it, where for a displacement $\vec{r}$, the electric field is given by:
$$
\vec{E}(\vec{r}) = k_e \cdot \frac{q}{r^2}\hat{r}
$$
The principle of superposition dictates that the total electric field at some point equals the vector sum of the electric fields of all the charges $q_i$ of displacement $\vec{r}_i$. That is,
$$
\vec{E} = k_e \sum_{i} \frac{q_i}{{r_i}^2}\hat{r}_i
$$
where $k_e$ is Coulomb's constant, $r$ is the distance from the charge, and $\hat{r}$ is the unit vector, directed away from $q$, such that another charge, $Q$, feels force
$$
\vec{F}_e = Q \cdot \vec{E}(\vec{r})
$$
For a positive charge, $q>0$, the electric force is in the direction parallel to that of the electric field. That is, $\vec{F}_e \parallel \vec{E}(\vec{r})$
For a negative charge $q<0$ the electric force is in the direction antiparallel to that of the electric field. That is, $\vec{F}_e \parallel \vec{E}(-\vec{r})$
Statement from Prof. Macdonald:
I don't care if you skip steps, when I see show you work I don't need you to write out every line of algebra you can do it in your head or you want to do it on the side like scratch paper for homework, go for it. You just need to show enough work that we can follow your process, so you need to show he steps. On exam probably don't want to waste time scribbling it off on the side, plus you want to make sure that nothing gets missed as far as partial credit.
Chapter 24 | Gauss's Law
Lecture 2022-07-05 ✅
Charge Density
- The charge density is the number of charges per unit length, area, or volume.
Length
For a continuous distribution of charge across a line of length $l$, the linear charge density $\lambda$ is defined in terms of charge per unit length ($\rm \frac{C}{m}$)
$$
\lambda = \frac{q}{A}
$$
The total charge $Q$ along $l$ in region $D$ is given by:
$$
Q = \int_{D}{\lambda ~ dl}
$$
Area
For a continuous distribution of charge across a surface of area $A$, the planar charge density $\sigma$ is defined in terms of charge per unit area ($\rm \frac{C}{m^2}$)
$$
\sigma = \frac{q}{A}
$$
The total charge $Q$ across $A$ in region $D$ is given by:
$$
Q = \int_{D}{\sigma ~ dA}
$$
Volume
For volume, the volumetric charge density $\rho$ is defined in terms of charge per unit volume ($\rm \frac{C}{m^3}$)
$$
\rho = \frac{q}{V}
$$
The total charge $Q$ within $V$ in region $D$ is given by:
$$
Q = \int_{D}{\rho ~ dV}
$$
The net flux through a Gaussian surface is given by:
$$
\Phi_E = \frac{q}{\varepsilon_0}
$$
And the net flux through any closed surface is given by:
$$
\Phi_E = \oint \vec{E} \cdot d \vec{A} = \frac{q_{in}}{\varepsilon_0}
$$
Where $\vec{E}$ is the electric field at any point on the surface, and $\q_{in}$ is the net charge inside the surface. Note that $\vec{E}$ includes both the charges inside and outside the surface.
The net flux through the spherical surface is proportional to the charge inside the surface. The flux is independent of the radius $r$ because the area of the spherical surface is proportional to $r^2$, whereas the electric field is proportional to $r^{-2}$. Therefore, in the product of area and electric field, the dependence on r cancels.
The net flux through any closed surface surrounding a point charge $q$ is given by $\frac{q}{\varepsilon_0}$ and is independent of the shape of that surface.