Why is voltage used when measuring the strength of a battery and not electric field?

Question

I'm not sure how to ask this but,

I started learning about electromagnetism with how charged particles behave in electric fields. This was easy and intuitive to understand. But I felt a disconnect when moving on to circuitry, as voltage, something I think of as potential energy per coulomb, is used as the main variable in equations rather than electric field.

It's easy for me to understand why/how electrons move in a wire due to an electric field on one end (with a source of charged particles) and a ground on the other end of a wire. However, when that is replaced by voltage, it becomes less clear conceptually for me and makes voltage just feel like a magic number representing the 'pressure' in a wire.

This is a bit difficult to explain, but in short what I think I'm trying to say is, very roughly, why is it $I=V/R$, and not $I=E/R$ (I know units for $R$ would have to change). Would it not be better to treat the electric field as the main driving force and not voltage (at least conceptually)?

I'm sure I'm missing something important and would appreciate any help in understanding the issue. Apologies if the question is unclear.

Answer 1

Would it not be better to treat the electric field as the main driving force and not voltage (at least conceptually)?

For circuits the use of voltage is far better than the use of E fields would be.

First, and most importantly, the main benefit of circuit theory is that the geometry of a circuit is not part of the theory. When you are drawing and analyzing the circuit for a house it doesn’t matter if you run the wire under the floor or through a wall. The circuit is the same. That abstraction requires the use of voltage. The E field is a vector so it has magnitude and direction and hence the geometry is essential. Also, different resistor materials will have different E fields for the same resistance, depending on the length. So in using the E field the geometry of the components would be required. Most of the simplifications afforded by circuit theory would be gone, and circuits would essentially require Maxwell’s equations to solve instead of Kirchoff’s laws.

Second, common electrical sources are voltage sources, not E field sources. Consider two batteries of different sizes but using the same chemistry. Those two batteries will have the same voltage but different E fields. You can change the orientation of the electrodes inside the battery, or the spacing between the terminals, or any number of other design details. Those changes will change the E field, but not the voltage.

I am sure there are other reasons, but these two are strong. Especially the first. I don’t think that modern electronics would be what they are if we insisted on designing them using Maxwell’s equations instead of Kirchoff’s laws.

Answer 2

The reason is just that voltage is easier to measure. It is all that you need to know to calculate currents in circuitry, but to express it via the electric field values you need to conduct a lot of experiments (in case of non-uniform field) and to sum up properly all this values. Why do that if you can just use voltmeter?

Answer 3

You certainly can (and do) consider the passage of charged particles around an electrical circuit in terms of electric fields but then you have to consider the practicalities.

You mighty start by saying that the work done by the electric field in moving charge $q$ from terminal $A$ to terminal $B$ within a battery (or resistor, circuit other elements) is $\displaystyle \int_{\rm A}^{\rm B} q\vec E\cdot d\vec s$.

But what does one do next?

Evaluation of the integral will usually be not that easy or practical as you need to specify the sign of the charge carriers and the electric field is a vector quantity whose magnitude and direction depends on position.

In the first instance it is easier to say that one does not really care as to what happens inside a circuit element but what is important is the work done in taking unit (positive) charge through a circuit element and call it $qV_{\rm AB}$ where $V_{\rm AB}$ is the potential difference which is a scalar and relatively easy to measure.

I do not know if you have covered Kirchhoff's voltage law?
You can certainly derive it from a consideration of the total work done when a charge moves around a circuit in terms of $\displaystyle \int_{\rm A}^{\rm B} q\vec E\cdot d\vec s$ type integrals but then rewrite it in terms of potential differences.

$\displaystyle \int_{\rm A}^{\rm B} q\vec E\cdot d\vec s + \displaystyle \int_{\rm B}^{\rm C} q\vec E\cdot d\vec s+ \displaystyle \int_{\rm C}^{\rm A} q\vec E\cdot d\vec s =0 \quad\Rightarrow V_{\rm AB} + V_{\rm BC}=V_{\rm CA}=0$

Answer 4

This is a bit difficult to explain, but in short what I think I'm trying to say is, very roughly, why is it $I=V/R$, and not $I=E/R$ (I know units for $R$ would have to change).

Because the electric field $E$ along a resistor is not the same as the voltage $V$ across the resistor. $E$ is the voltage gradient along the resistor. Assuming a resistor of length $L$ with uniform resistivity and cross sectional area so that the electric field would be relatively uniform in the resistor, then the electric field would be the voltage across it divided by its length, or $E=V/L$. So if you wanted to express the current in such a resistor in terms of the electric field it would be $I=V/RL$

Would it not be better to treat the electric field as the main driving force and not voltage (at least conceptually)?

While you certainly can, in circuit analysis it would not be practical (at least in my opinion). Though the electric field is the driving force behind the current, of interest is the voltage across the resistor (work per unit charge required to move the charge between the end points) since that is what is used for the application of Kirchhoff's laws in circuit analysis.

Hope this helps.