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

field of a negative point charge

(Courtesy of wikipedia)

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Small note: "induction" has a specific meaning in electromagnetism having to do with the effect of having other materials present. For this question you can just say "the electric field of a charge." – Michael Brown Mar 20 '13 at 14:01
up vote 4 down vote accepted

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.

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I think the equation is: F=k (q_(1 ) q_2)/r^2 – Samama Fahim Mar 20 '13 at 14:22
@SamamaFahim Thanks; I fixed the powers. – joshphysics Mar 20 '13 at 14:25

There's a few ways to answer this. The best way to understand this is that it's true simply by the construction of electromagnetic fields, and field theory. In general, a field will always point from higher to lower potential energy. This is represented by the equation: $$\vec{E} =-\nabla V $$

This means that the electric field is the negative gradient of the electric field. The reason it is constructed this way is because any object in a potential energy field will accelerate towards a lower potential, so using the above equation, the field lines point in the direction of the force.

A positive electrical charge has more potential energy the farther it is from the negative charge. As it gets closer, it loses potential energy (which becomes kinetic energy). This means that the closer you get to the negative charge, the less potential energy it has.


Since $q_1$ is negative, the potential energy decreases (becomes more negative) as r decreases. So in conclusion, just think of an electric field, or any field, as pointing in the direction of decreasing potential energy. This is simply the construction of fields.

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If I think of an electric field, or any field, as pointing in the direction of decreasing potential energy, then if we bring a positive test charge near a source charge which is positive as well, then in bringing them closer the potential energy will increase and the electric field points towards decreasing potential energy. What if we pull away a negative charge from a positive one then its potential energy will increase in the direction where it is supposed to be decreasing in case of a positive test charge? – Samama Fahim Mar 20 '13 at 15:18
@SamamaFahim Yes, bringing a positive test charge closer to a positive source charge will increase its potential energy, so the electric field of a positive source charge points away from the charge. Yes, pulling a negative source charge away from a positive one would increase its potential energy. This seems to run counter to what I said above, but I used the example of a positive source charge for a reason. There is a difference between electrical potential and electrical potential energy. – Ataraxia Mar 20 '13 at 15:39
@SamamaFahim Electrical potential energy can be calculated from electrical potential with the equation $U=qV$. What the field really corresponds to is the direction of decreasing electrical potential, not necessarily electrical potential energy. When the test charge is positive, the electrical potential energy has the same sign as the potential. When it is a negative test charge, the electrical potential energy has the opposite sign. Does that clarify it? – Ataraxia Mar 20 '13 at 15:43

According to me positive charge is present in nucleus so it will attract towards negative charge and -charge is present outward nucleus so attraction is radially outward.

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