How do electric field lines show how the electric field mediates force?

Question

I had a thought the other day about electric field (and electric field lines) and how they show repulsion/attraction between corresponding charges respectivelly. Some places I’ve read talk about how the “lines” contract length wise (attraction) and apply lateral pressure transversally (repulsion). But my confusion (and where my question comes from) is if electric field lines are thought to “contract” then why do they radiate outwards from their charge?

I understand this can be thought of as just an analogy, but what would help close this gap for this analogy? Thanks!

Answer 1

We could think only of forces at a distance between charges, as stated by Coulomb. But all the electromagnetism was built in the $19^{th}$ century around the notion of fields.

So, instead of saying that a negative test charge ($-q$) is attracted by a positive charge ($+Q$) with a force $\mathbf F$, we postulate a vector field $\mathbf E$ around ($+Q$), so that for each point $\mathbf r$ there is an $\mathbf E(\mathbf r)$. When ($-q$) is at the point $\mathbf r_0$, it feels a force $\mathbf F = -q\mathbf E(\mathbf r_0)$ from the field. By convention, for a positive source ($+Q$), the arrows of the vector field points outward. As the force for a negative test charge ($-q$) has a negative sign, it points inward.

Following the idea, all combinations of positive or negative sources, and positive and negative test charges result in the appropriate force vector.