Drawing field lines

In problem 16, you sketched the vector field $\myv v= \frac{1}{\rr^2} \uv \rr$. The amplitude $|v|$ drops off as $1/r^2$ away from the origin, which most of you captured quite nicely with shorter arrows at greater distances.

This differs only by a constant from the electric field of a point charge $\myv E = \frac{q}{4\pi \epsilon_0}\frac{1}{\rr^2} \uv \rr$.

This way of drawing field lines was rather cluttered, so instead let's just draw connected lines in the local direction of the electric field.

Does that mean that we've lost the information related to the magnitude of the field?

Actually, that info is still preserved by considering that the density of field lines $\propto$ to the field strength.

So that in the orange region, the field lines are closer together $\Rightarrow$ the field has a greater magnitude in the orange region compared to the blue region, where the lines are less dense.

(Actually, we need to think in three dimensions: the field is dropping as $1/r^2$, and so the "density of field lines" metaphor only holds if we think of field lines spread out over a 3-d spherical surface, which has an area $a=4\pi r^2$, instead of a 2-d circle with circumference $c=2\pi r$.)

Further rules for drawing field lines

• The number of field lines emanating from (or ending on) a charge should be proportional to the charge.
• Close to any single charge, the lines should reflect the short-range radial symmetry around the charge (shown: two equal and opposite charges)
  
• $\myv E$ lines point away from positive charges, and towards negative charges.
  
• Indeed, field lines may never cross (How could $\myv E$ point in two different directions at the same time?),
• and may only ever start or end on some chunk of charge.
  
  

Indeed, as intuited by our plumbing models, these rules suggest positive charges as sources or "faucets" of water flow electric field lines, and negative charges as sinks or "drains" for electric field lines.

1. Imagine 2 1-meter long segments, one with a total charge of +8 C and the other with a charge of -8 C. The charge on each segment is evenly distributed. Sketch the field lines when the two segments of line charges are 1 km apart.
2. Now, imagine that the two segments are just 1 mm apart from each other (and parallel). Sketch the field lines in the region (say about 5 mm on a side) close to the middle of the whole setup.

We shall further, define another very plumbing-like quantity which is the flux.

$$\Phi_E = \int_S \myv E \cdot d \myv a.$$

This is something like the "flow" of electric field vectors through the surface ${\cal S}$. (It corresponds to the fluid mass / unit time crossing the surface, if the vectors represent fluid velocity instead of electric field.)

Imagine that a charge is positioned at one corner of a cube, as shown. The flux of electrical field through a surface, $S$ is given by: $$\Phi_E=\int_S \myv E\cdot d\myv a.$$ It turns out that $\Phi_E=0$ for the turquoise-colored surface shown. Explain why this is so.