September 20 2026

Today was a really busy day - I had church and violin again, and I spent a bit of the night studying some of Purcell Chapter 2. Multivariable Calculus isn’t as terrifying as I thought, it’s actually quite interesting! (This sounds so nerdy, I know)

Currently, I’m thinking of just making a notes tab to put my notes per chapter in for easy access. I’ll also start my Further Maths/Personal project grind tomorrow and try to not waste time once I come home.

Here’s a quick summary of what I learned about E&M over the past 2 days:

Electricity and Magnetism Chapter 2: Part 1.

Potential

Electric Potential is defined as:

$\phi = -\int \mathbf{E} \cdot d\mathbf{s}.$

It has the units of Joules/Coulomb. Note that it’s different to potential energy as it uses a field instead of a force.

The potential difference is the difference in potential between two points. Often, we fix one of the points at infinity. However, this doesn’t always work.

For the electric force, this potential is path-independent.

The Gradient

The gradient turns a scalar function into a vector one in the following way:

$\nabla f(x,y,z) = \frac{\partial f}{\partial x} \hat{\mathbf{x}} + \frac{\partial f}{\partial y} \hat{\mathbf{y}} + \frac{\partial f}{\partial z} \hat{\mathbf{z}}.$

We know that:

$d\phi = -\mathbf{E} \cdot d\mathbf{s}.$

We also know that:

$d\phi = \frac{\partial \phi}{\partial x} dx \hat{\mathbf{x}} + \frac{\partial \phi}{\partial y} dy \hat{\mathbf{y}} + \frac{\partial \phi}{\partial z} dz \hat{\mathbf{z}}.$

and that:

$d\mathbf{s} = dx \hat{\mathbf{x}} + dy \hat{\mathbf{y}} + dz \hat{\mathbf{z}}.$

Therefore, we can infer that $\boxed{\mathbf{E} = -\nabla \phi}$.

Energy in a system using potential

I’m just gonna list the equation cuz it’s pretty self-explanatory:

$$U = \int_V \rho \phi dv.$$

Basic Dipoles

Chapter 2 of Purcell+Morin has a brief introduction to dipoles, explained with an approximation that the distance from the dipole that we are looking from is far greater than the distance between the two poles.

It can be shown that (proof in the textbook): $\phi = \frac{ql\cos{\theta}}{4\pi \epsilon_0 r^2}.$

Note that we used polar coordinates for simplicity, where $\theta$ is 0 from the top and increases clockwise.

I’ve also learned the shapes of the field lines for the potetial + electric field for the dipole, but not really that well. I’ve also learned a bit of the divergence + gradient in polar coordinate, but the detailed study of that’s for tomorrow me! I’m gonna go sleep now, it’s 10:05 pm and I’m very happy to be finally able to sleep early for once and overcome the mental barrier of sleep lol, goodnight!