electrostatics - Electric field of a negative charge

How was it discovered that the electric field of a negative charge points towards the charge itself? Is it true?

(Courtesy of wikipedia)

Answer

The electric field of a negative point charge points towards the point charge as a result of the definition of the electric field of a point charge. To see this, recall that the electric field of a point charge $q$ is defined as $$ \mathbf E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\mathbf e_r $$ where, $r$ is the distance to the charge, and $\mathbf e_r$ is the outward pointing radial unit vector field emanating from the location of the charge. If $q$ is negative, then we see that the electric field points in the direction of $-\mathbf e_r$, and therefore it points radially inward.

You might then ask "well this is fine, but why is the electric field defined in this way? Couldn't we have defined the electric field in such a way that the electric field due to a negative point charge points radially outward?"

Well, the definition of the electric field is motivated by Coulomb's Law, the empirical fact that the electrostatic force exerted by a point charge $q_1$ on a point charge $q_2$ is $$ \mathbf F_{12} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\mathbf e_{12} $$ where $\mathbf e_{12}$ is the unit vector pointing from charge 1 to charge 2.

The idea in defining the electric field is that we want to associate a vector field (which we call the electric field) to each point charge individually such that if we multiply that field by any other charge (which is usually called a test charge), then we obtain the force that would be exerted on the test charge due to the original charge. If we factor $q_2$ out of the right hand side of Coulomb's law, then the rest of the stuff is precisely such a vector field associated to the charge $q_1$ $$ \mathbf F_{12} = q_2\underbrace{\left(\frac{1}{4\pi\epsilon_0}\frac{q_1}{r^2}\mathbf e_{12} \right)}_{\text{stuff that only depends on charge 1}} $$ so we define the electric field as the stuff in parentheses. Notice, however, that we could just as well have factored out $-q_2$ in which case the electric field would have been opposite in sign to the conventional definition, and in this case, the electric field of the negative charge would have pointed radially outward by definition.