# Electric Field support by two negative or positive charges

Is it possible for negative charges to ride along the electric field lines going in to another negative electric charge? For instance say we have two negative charges of $-5C$ and $-2C$, clearly the electric field near the $-2C$ is stronger than the electric field lines near $-5C$, so when $-5C$ moves nearer to $-2C$, can $-5C$ charge ride along the stronger field lines near the $-2C$ charge?

• You should look into Coulomb's law. Commented Oct 23, 2021 at 14:30

Is it possible for negative charges to ride along the electric field lines going in to another negative electric charge?

It depends.

If both charges are point charges, all their field lines will go to infinity, i.e., there won't be any lines connecting one with the other. This is because field lines, originating on a negative charge, either terminate on a positive charge or go to infinity.

If the charges are actually too negatively charged finite size conductors (charged bodies), the situation will depend on the distance between them.

If the bodies are far away from each other, all their field lines will go to infinity - just like if they were point charges.

But, if the bodies are getting closer to each other, their charges get redistributed and their fields will change. In particular, since both bodies are negatively charged, they will push each other's negative charges to the outside surfaces (surfaces of the two bodies farther away from each other), leaving fewer negative charges on the inside surfaces (surfaces of the two bodies facing each other).

At some point, when the bodies are close enough to each other, all excessive negative charges of the $-2C$ body will migrate from the inside surface to the outside surface and, after that, negative charges leaving the inside surface will be creating excessive positive charges on the inside surface.

As soon as this happens, some of the negative field line originating at the inside surface of the $-5C$ body, will connect to the newly created positive charges on the inside surface of the $-2C$ body. Now, a negative charge could ride one of these lines, all the way from the $-5C$ body to the $-2C$ body.

Yes, but I think not in the way you imagine. The $-5\ C$ charge will "ride" along the field lines near the $-2\ C$ charge, but away from that charge. At the same time, the $-2\ C$ charge will "ride" along the field lines near the $-5\ C$ charge, also away from that charge (assuming both are in free space and are not constrained in some way from moving).

The force between electric charges is described by $F=\frac{kq_1q_2}{r^2}$. If the force $F$ is positive, because the charges are like charges (both positive or both negative), then the force is repellant, or if the charges are unlike charges, (one positive and one negative) then the force is attractive.

Similarly, the electric field strength is given by $E=\frac{kq}{r^2}$. The similarity is that the strength of the E-field and force are governed by the inverse square law.

There is one assumption in the question that is not right: that the E-field near the $-2\ C$ charge is stronger than the E-field near the $-5\ C$ charge (assuming that the distance from the charge is the same). In fact, at any given distance, the E-field near the $-5\ C$ charge will be $2.5\times$ the E-field near the $-2\ C$ charge.

The sign of the charge represents whether the charge is positive or negative, because it has implications with two charges (whether the charges are like or unlike). The magnitude of the E-field or force ($|E|$ or $|F|$), or the strength of the field, is absolute. Thus $|\frac{k(-2)}{r^2}|$ is less than $|\frac{k(-5)}{r^2}|$ just as $|\frac{k(-2)}{r^2}|$ is less than $|\frac{k(+5)}{r^2}|$, and $|\frac{k(+2)}{r^2}|$ is less than $|\frac{k(-5)}{r^2}|$.