# Electric voltage versus gravitational voltage across a unifom field

Let us say we have a uniform electric field, like between two charged plates separated by a distance $$d$$.

The formula for the voltage between the plates is $$\Delta V=Ed$$.

But what is the value of this uniform field $$E$$? When I look for the formula, I see $$E=\Delta V/d$$, but this circular reference doesn't help. Well, when the field is not uniform, the formula is $$E=kQ/r^2$$, but what is $$r$$ here?

With gravity, it is easier. For example, one measures that the gravitational field close to the surface of the Earth (ultimately acceleration) is $$g$$ = $$GM_{Earth}/R^2$$ and you assume that this is not going to change significantly if you add to that whatever height, which is negligible compared to the radius of the Earth $$R$$. Then the gravitational potential difference, sometimes called "liftage", is the same thing over an additional distance $$h$$, i.e. $$\Delta V=gh$$. In conclusion, liftage is proportional to the constant $$G$$, the source mass $$M_{Earth}$$ and the height $$h$$ and inversely proportional to $$R^2$$.

My problem with the electric potential is that I only see the additional distance $$d$$, not the original distance $$r$$. So what is electric potential or voltage dependent on, is it only proportional to the constant $$k$$, the source charge $$Q$$ and the distance $$d$$?

• Your "not uniform field" is actually just true for point charges (or outside of symmetric charge distributions). It's not an equation for all non-uniform static electric fields. Dec 16, 2019 at 15:16

The treatment of the field between two thin, infinitely large, plates with equal but opposite surface charge densities can be found in any good introductory physics/E&M text.

You can use Gauss's law (work left to you) to show that the field due to one plate is given by $$E=\sigma/2\epsilon_0$$, where $$\sigma$$ is the surface charge density on the plate, and the field is oriented perpendicular to the plate.

For two parallel plates with equal but opposite surface charge densities, we can easily add the fields together and get the field between the plates to be $$E=\sigma/\epsilon_0$$ pointing from the positive plate to the negative plate, and $$E=0$$ outside of the plates.

This relates to the equation you give of $$\Delta V=Ed$$ because this is the definition of potential difference between two points in a uniform field, where $$d$$ is the component of the displacement between the points in the direction of the field. Going further to something you might have seen before, this means that $$\Delta V=d\sigma/\epsilon_0$$, and therefore the capacitance of a parallel plate capacitor is $$C=Q/V=\sigma A/V=A/d\epsilon_0$$, where $$A$$ is the area of one of the plates.

The reason you might see your "circular" equation $$E=\Delta V/d$$ is because typically "in the lab" we directly control what the potential difference is between the plates (and their separation), and hence this equation is more useful from a practical/interpretive(?) point of view, even if the equations are mathematically equivalent.

I will note that you seem to make a distinction between gravity and electrostatics, but mathematically they are built up from the same form, since they both start with fields $$\mathbf F$$ from point particles of the form $$\mathbf F=\frac{k}{r^2}\hat r$$ In other words, if we had a planet with a large net charge (ignoring the interaction of charges within this planet, and ignoring gravity (not the most valid physical assumptions, but that's not the point here)) and we threw an oppositely charged ball near the surface of this planet, we would see exactly the same behavior as if we threw a normal ball here on Earth. In the same way, if we somehow had a large, thin, massive plate and a large, thin, "negatively massive" plate (once again, not physically real), then we would expect this to behave just like our parallel plate capacitor.

So, if you think gravity is "easy" then electrostatics should be too :)

• Thanks! I was precisely trying to unify my notes on gravity and electrostatics and forgot the distinction btw point masses/charges vs. extended objects, so I mixed apples (point masses) with pears (capacitor). I should have compared $g$ with $E$ caused by a charged sphere with homogeneous charge distribution, right? And in a 2nd table for extended objects compare "mass capacitor" with "charge capacitor", as you did, but why do you need the second massive plate to be unrealistically negative? You only need the plates to attract each other and for mass that happens without signs. Dec 16, 2019 at 20:50

I don't know if you know calculus, but based on your question and the use of $$\Delta$$ quantities, I suspect you don't.

The electric field due to a point charge,$$Q$$, is, as you say, $$\vec{E}=\frac{kQ}{r^2}\hat{r},$$ where $$r$$ is the distance from the charge to the point of interest in space, and $$\hat{r}$$ is a unit vector pointing from the charge to the point of interest.

If you want the electric field at a point in space due to a collection of charges (a line of charge, a plane of charge, a ring of charge), you must integrate the individual field contributions over the total collection of charge. When you do that for a plane of charge, uniformly distributed, you get the field which is approximately constant for distances $$d$$, from the plane which are small compared to the size of the plate. The voltage (aka, change in potential) between the plates will change linearly because the electric field is the negative slope of the potential). If the slope is constant, the function is linear.

When you do the integral to find the electric field and apply that to a conducting plate, the resulting magnitude is $$E=\frac{Q}{\epsilon_0 A},$$ where $$\epsilon_0$$ is the permittivity of free space.

• Very helpful. I have the basics of calculus... As I was thinking of a uniform field (constant force), I thought that calculus was not needed. But I was also thinking of point charges, mixing them up unduly with an extended object like a capacitor. I understand that, in the latter case, calculus (integration) is needed to compute the contributions of each element, unlike what would happen in a sphere with equal charge distribution, which can be treated as a point charge, right? I also assume that your formula is the same as Aaron's because $Q/A$ is charge density. Dec 16, 2019 at 21:00