# Why does field strength follow the inverse square law but potential does not?

Either in a gravitational or electrical field, let's say an electrical field, the electrical field strength follows the inverse square law. This is fairly intuitive just due to the geometry of the field. However, the electrical potential of a charged particle at a point in the field does not follow the inverse square law. Why is this?

Context:

I am studying Physics at A-Level (UK qualifications, exams taken at 18 years old). The equations are often presented just to be remembered, with little exploration of where they come from, so I'm trying to get a proper understanding of these equations.

• Before writing an answer, I would like to ask this: why are you surprised that a scalar field's dependence on distance is different from dependence on distance of that field's gradient? – Alfred Centauri Dec 22 '19 at 22:21

In the A-Level, you are already working with single variable calculus; if you are working in multivariable calculus and know what a gradient is, this will make more sense, so please comment if you do.

The correct equation to consult here would be the relationship between a force and a potential: energy:

$$F = -\frac{dW}{dx}$$

Intuition: A field exerts a force on a particle on the direction of maximum decrease of energy (because, empirically, we found that the universe wants everything to have the least potential energy it possibly can, at least in classical physics). In the single variable case this “direction” can be either $$+x$$ or $$-x$$, however, in the case where we have more than 1 dimension, this is much more illuminating: suppose you place a ball on a hill.Where will it go? You will find it accelerates (I.e feels a force) in the direction of steepest descent. This thinking applies analogously to electrostatics.

For a given particle, the relationship between electric potential energy and the electric potential is $$W = q V(x)$$ where $$q$$ is the charge of the particle, $$W$$ is the potential energy and $$V$$ is the potential. This one is less fundamental: The potential is a property of the field, but two particles in the same $$x$$ would have the same potential but different energies, depending on their charge.

For completeness, the relationship between a force exerted by the electric field on the particle, and a electric field, is $$F= qE(x)$$ Because, in similar fashion to before, two particles with different charges in the same point in space should feel a force proportional to their charge.

With these three equations in mind, we should see that the relationship between an electric field and a electric potential is a derivative:

$$E = -\frac{dV}{dx}$$ and thus an inverse square force/electric field should correspond to an inverse potential.

• I still can't get my head around this. If the electric field is causing the particle to have potential energy, and the strength of that field follows the inverse square law, surely the potential energy should follow the inverse square law too. I'm also thinking of the equipotentials drawn on a diagram of a radial field, they get further apart the further from the field you look. Is that not the inverse square law? – zacccczn Dec 22 '19 at 19:54
• What you have written here is correct. I think maybe what is actually confusing you is overloaded terminology. There is a potential energy and also just a potential. – JohnA. Dec 22 '19 at 19:55
• Yeah, it's very likely I'm just lost in my own confusion. I'll have to come back to it tomorrow, some sleep and time away from thinking about it always seems to help. Thanks for the responses. – zacccczn Dec 22 '19 at 20:00
• @zacccczn This is an appropriate answer. The equipotential surface you are talking about will come from any inverse relation not necessarily inverse square only. From $E(x)=-{dV\over dx}$, you can show that electrical potential will always follow inverse law. – Sam Dec 22 '19 at 23:27

One can look at it as follows: The field has to fall off as $$1/r^2$$ because the flux through a spherical surface is independent of distance, and of course, the force is proportional to the field:

$$\vec F = q\vec E$$

The energy required to get a charge from $$\infty$$ to $$r$$ is computed from the work done along the path:

$$W(r) = \int_{\infty}^r{F(r')dr'}= \int_{\infty}^r{\frac{q}{r'^2}dr'}=-\frac q r$$

which is also called the potential.

• You've confused potential energy for the potential (I mixed the two up as well at first). If you're talking about potential energy, then it should be $W(r) = \int Fdr' = \int q_{1}q_{2}/r'^{2} dr' = -q_{1}q_{2}/r$ (you must have two charges). If you're talking about the potential, then it doesn't have to do with work (not directly at least). The potential is the (negative of the) line integral of $\vec{E}\cdot\vec{r}$, which is $q/r$. – Maximal Ideal Dec 22 '19 at 20:44
• the whole answer is per unit charge, since the point is whence does the $r$ dependence originate. The $q$ is the fiducial test charge, and whatever make $\vec E$ doesn't really matter. – JEB Dec 23 '19 at 5:32