# How does the electric field produced by a simple circuit look?

I have not seen anywhere a description of how the electric field looks inside and around a simple circuit. For example let's say we have the circuit shown below. One DC voltage source, two resistors, and a constant current flowing around.

We know that the electric field inside the battery will point from positive to negative, we also know that the field inside the wires is very small and in the direction of the current. Through the resistors there will be a strong field pointing from positive to negative. But in order to maintain the relationship that a closed loop integral of the E field is zero everywhere we must also have a field outside of those circuit elements. I have no idea how this field will look but I have made a crude attempt at sketching it below.

Is this a realistic picture of how the field will look?

• As a first approximation, looks good to me. – V.F. Jul 4 '18 at 21:48
• As others on this site have pointed out there is a book Matter & Interactions by Chabay and Sherwood which goes into this in some detail. See also their paper A unified treatment of electrostatics and circuits. – nadder Jul 11 '18 at 20:10

An electrical circuit is a lumped element model, that does not carry any geometrical information with it. Given a circuit you do not know what is the shape of the resistors, the dimension of the battery, the cross section of the wires, the position of the elements with respect to each other, and so on and so forth. Unless, you know those information by other ways, there is no way to even sketch the electric field.

Moreover, the idealized wires that connects the various components of the circuit are mere topological connections, they do not have any electric field inside.

a closed loop integral of the E field is zero everywhere

That is not what the second Kirchhoff's law says. It can easily be derived from the Faraday's law under the assumption that the magnetic field does not change.

$$\oint_{c} E \cdot \text{d}l=-\int_S \frac{\partial B}{\partial t}\cdot \text{d}S$$

Since, $B$ does not change with time its derivative is zero , thus

$$\oint_{c} E \cdot \text{d}l=0 \;\;\;\;\;\;\;\;\;\;\;\; \forall c$$

it must be true for all possible close loops c. This equation does not state anything about the field out of loop. Simply, it says that the sum of the electric field all around a loop is zero.

• I don't think I agree with this answer. I think you can certainly sketch the $\vec{E}$ field for the circuit even without knowing all the element information. You can still know, for example, the electric field is greater inside the resistor than outside. You can't just calculate the numerical magnitude without more information. – Zack Hutchens Jul 8 '18 at 13:06
• @zhutchens1 Although, you may say that the electric field is nonzero inside the elements and zero elsewhere. I think it is meaningless because in an electric circuit the elements are ideally concentrated in a single point, the whole idea of $E$ losses its significance being the electric field a distributed property of space. Anyway, certainly the sketch with the red arrows in the questions is meaningless. – Dante Jul 8 '18 at 13:18
• @Dante I see what you mean, maybe I should have built a circuit out of real components and taken a picture of that instead so that it's more clear that I'm considering a real circuit with non-ideal components. Of course it's impossible to exactly sketch the field if you don't know the exact makeup of all components but I'm more interested in an approximate qualitative description. About Kirchoff 2, I don't see the difference between your statement and mine, what you wrote is what I tried to say anyway. – nadder Jul 8 '18 at 14:19
• @nadder ok, it's a very reasonable question, although I don't think it is well stated if that's what you wanted to know. Basically, because your question can't be addressed in the framework of the electrical circuits theory. You should have written something like: given the electric field in this loop, how does $E$ look like elsewhere, knowing that Kirchhoff's second law holds? – Dante Jul 8 '18 at 20:03