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

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!

• Are you familiar with the "Maxwell stress tensor"? Commented Dec 26, 2022 at 21:24
• I am not but it sounds like it may help me bridge the gap in my misunderstanding.. could you explain it in laymans terms?
– Jake
Commented Dec 26, 2022 at 21:29
• It is hard to do that without knowing your background in electrodynamics. It is a tensor being bilinear in the components of the electric and magnetic field that allows you determine the forces exerted by the electromagnetic field acting on any spatial region of your choice. In particular, in the case of two charges you see immediately how attraction/repulsion is related to the form of the field lines. You find this in all good textbooks on theoretical electrodynamics (like Landau/Lifshitz vol. 2 or Sexl/Urbantke, Relativity, Groups, Particles, etc.) and probably also in Wikipedia. Commented Dec 26, 2022 at 21:50
• @Hyperon i only have an undergraduate physics class experience (I was an engineering major) so I have never learned much more beyond that
– Jake
Commented Dec 27, 2022 at 1:02

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.