Suppose you have a two wires. Each has a current of $5 \text{ A}$. One is made of copper (Cu) and has radius $0.5 \text{ mm}$. The other is made of zinc (Zn) and has radius $0.1 \text{ mm}$. What is the drift speed of electrons in each wire? You may want to consult the table below.

• The copper wire has $I=5 \text{ A}$, $r = 0.5 \text{ mm}$.
• The zinc wire has $I=5 \text{ A}$, $r = 0.1 \text{ mm}$.
• The charge of an electron is $q=-1.6\cdot 10^{-19} \text{ C}$.

• Drift speed for both wires.
• Electron charge density for both wires.
• Electron current for both wires.

• The table is accurate for our wires.
• The wires have circular cross-sections.
• The wires do not experience any external electric field.

• We represent the situation with the following diagram.

First, notice that we probably do not want to do any calculations here, since the it will not be fun to take a dot-product of the dipole's electric field and the area-vector, and it will get very messy very quickly when we start integrating over the surface of the cylinder. Instead, we evaluate the situation more qualitatively. Consider the electric field vectors of the dipole near the surface of the cylinder:

Notice that the vectors near the positive charge are leaving the cylinder, and the vectors near the negative charge are entering. Not only this, but they are mirror images of each other. Wherever an electric field vector points out of the cylinder on the right side, there is another electric field vector on the left that is pointing into the cylinder at the same angle. These mirror image vectors also have the same magnitude, though it is a little tougher to visualize.

We could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored for left vs. right, and the $\vec{E}$-vectors are mirrored, but opposite $\left(\vec{E}_{left}=-\vec{E}_{right}\right)$. We tentatively write the equality, $$\Phi_{left}=-\Phi_{right}$$

Putting it together, we tentatively write: $$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$ We gain more confidence when we read the next section of notes, where we define “Gauss' Law”. This law states that the total flux through a close surface is the amount of charge enclosed divided by $\epsilon_0$, the permittivity of free space. $$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ Since the total charge of the dipole is $0$, then indeed the charge enclosed is $0$, and we were correct with our reasoning about the electric field and flux above.