Is there a small spark every time a switch is closed to engage a circuit?

Question

Suppose we have a simple circuit with a resistor $R$ connected to a DC voltage source $V_{text{batt}}$. The circuit can be enabled or disabled by a switch. The dielectric breakdown of air is $E_{text{breakdown}} = 3times 10^6 text{ V/m}$. This value will be exceeded when the switch contact is a distance
$$
d = V_{text{batt}} / E_{text{breakdown}}
$$
If $V_{text{batt}}$ is small then $d$ will also be small, but there will always be some small distance where the breakdown will be exceeded and I would expect a spark.
If $V_{text{batt}}$ is 10 V then $d = 3.3 hspace{2 pt}mutext{m}$. This seems like you could notice a spark at this distance?
So is it correct that there is always a small spark when a switch closes? If not why?
I could imagine entering a regime with small $V$ where $d$ is so small such that $tau = d/v$ where $v$ is the switch contact velocity is small enough that there isn't really time for the air to break down or the spark would be so short lived that it's not really measurable in any way.
This question inspired by If I touch the “+” side of a high voltage battery, will I get shocked?

Note: I know that there is often a spark due to inductive loading when a switch is opened. This is NOT the effect I'm interested in. I'm interested in what happens when a switch is closed.

niels nielsen · Accepted Answer

To enlarge slightly upon DKNguyen's answer:
Indeed, the thing that causes air to break down is the strength of the electric field in the gap between the switch terminals, as measured in volts per meter. For small values of the gap, even a small number of volts will produce a strong field.
Furthermore, if there are any sharp points or asperities on the contact surfaces, the field will be locally magnified at those sites, and dielectric breakdown will be initiated there.
However, also note that the kinetics of dielectric breakdown of gases have time delays associated with them, as they rely on statistics to provide triggers for ion production. It is common in experimentation with deliberate spark formation in gaps to not get a spark when your measurement tool is gated open simply because the gate time wasn't sufficient to accomodate the random perturbations required to trigger the discharge. For this reason, ultraviolet light will be beamed into the gap to promptly produce photoelectrons at the start of each test shot so you always capture a nice spark on every shot.
Absent that UV, it isn't assured that you'll get a spark every time you turn on the electricity.
If you want repeatable sparks, they can be encouraged by filling the gap with a specific gas chosen for its breakdown characteristics. In applications where sparking was desired (by, for example, in so-called spark transmitters), a variety of techniques including special gas atmospheres were used to stabilize and make uniform the sparks which were used to excite radio waves prior to the invention of vacuum tube circuits.
Switches can be deliberately designed to be "bounceless" by pushing the contact surfaces together with sliding motion instead of snapping them together under pressure, and by furnishing multiple slider contacts in parallel, in which the first slider to make clean contact short circuits the other sliders, and if that contact happens to bounce or rub against a chunk of dirt then the redundant sliders in the array pick up the current flow without interruption.

DKNguyen · Answer

I assume you're asking about a mechanical switch and not a solid-state switch. Yes, there is a spark when you flip a mechanical switch closed because doesn't just close cleanly. It closes, then bounces open, then closed, then bounces open, etc. until it finally stays closed.
The voltage producing the arc is not the voltage from the supply. It is the voltage produced by the inductance when current flowing through it is suddenly interrupted and causes the magnetic field to collapse, releasing the energy stored in the magnetic field to try and maintain the current flow through the inductor prior to the interruption as best as it can. This voltage, is dependent on the inductance and the current running through it (i.e. the energy in the inductor) and can be very high even if the supply voltage is low.

Is there a small spark every time a switch is closed to engage a circuit?

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