# How can we assume that we're in electrostatics if we are talking about current?

In the textbook I am currently reading it states that to even out the flow of current throughout a circuit, the force required is of the form $f=f_s+E$ where $f_s$ is from the battery or other energy source. It then states that the emf $\mathscr E=\oint f \cdot dl=\oint f_s \cdot dl$ since $\oint E \cdot dl=0$ in electrostatics. My question is how can we assume that we're in electrostatics if we are talking about current? Wouldn't that just contradict one another?

• Please define your variables. If $f$ is a force, how does it have the same dimensions as $E$ which I presume is an electric field? – DanielSank Jan 11 '15 at 19:09

Electrostatics refers to slow-varying fields with time (i.e. constant fields) and it is not in contradiction with particle movement.

Think about a constant field acting upon an electron, we would be in the electrostatic regime (since the field is constant) and yet the electron would move.

Electrostatic or electrodynamic only refers to the time evolution of the fields. If you're referring to currents in electrical circuits, there's indeed a time variation of the field, but it is such that its wavelength is much larger than the typical size of the circuit, therefore electrons "feel" a constant (static) field and their velocity is the same, making possible for us to define a current.

Note: Particle movement must be slow too so that electrical charges would not radiate fast oscillating fields, but this is perfectly fulfilled in a circuit since the drift velocity of electrons is really small.

• Welcome to Physics! Note that we have MathJax embedded (effectively Latex) in the site for use in equations. – Kyle Kanos Jan 11 '15 at 18:31
• @manuel91: Will the curl of a changing electric field be non-zero? – Sidd Jan 11 '15 at 19:11
• It could be, since the curl operator acts only on spatial coordinates. Imagine a field that is constant in space but oscillates in time, such as the one generated by a paralell plate capacitor, the curl would be zero since the field would only change temporally and not spatially, therefore the spatial derivatives in the curl would be zero. – manuel91 Jan 11 '15 at 19:28
• @Sidd The curl has nothing to do with time... Not sure if you question is about Maxwell equations. – TZDZ Jan 11 '15 at 19:29
• @Sidd The line integral around a closed loop is proportional to the area of the loop and to the average value of the part of the curl orthogonal to the area enclosed. And the path is through space only, all at one time. – Timaeus Jan 11 '15 at 19:39

If you hook up a battery, then at first electrons might rush in one side (and leave a charge imbalance in the wire they came from) and electrons might rush out of one side (an thus leave a charge imbalance in the wire they entered).

But these charge imbalances cause electrons to move into or away from these regions. Even though the conduction electrons move slowly, these charge imbalances even themselves out very very quickly. The solution after the charge imbalances are evened out is what you are trying to solve for, not the transitory very short lived phase where you have to take into account that parts of the circuit more than $t/c$ units away don't yet know about the battery being plugged in. When you have to take into account the finite lag of information, that's when you aren't doing electrostatics anymore.

If you have a circuit that is small (in physical size) relative to ($c$ times) how quickly things change, then you can often use the electrostatics approximation to get accurate results.

If you have a steady current then you can also often avoid a changing magnetic field, which is also important for getting $\oint \vec{E}\cdot d\vec{l}=0$. That might not be true in the brief period of time where you go from no current to the final steady current (the degree to that violation is proportional to the self-inductance of the circuit).

Electrostatics refers to that class of electrical problems where the time derivative of all quantities is zero. You can have a current as long as it doesn't change with time.