Electric Potential and Corona Discharge

Question

I saw a physics demonstration where, between two flat metal plates, was a round conducting ball with large radius. When the two plates had a potential difference, there was an increasing electric field, which increased the electric potential of the conducting ball according to the equation $V = Er$, where $V$ is the electric potential, $E$ is the vector quantity of the electric field, and $r$ is the radius of the sphere. The large electric potential caused a spark across the air.

Then, a spike was placed between the two plates. Electric potential built up in the spike, but because the radius of the "ball" at the end of the spike is small, the electric potential at the end of the spike is small too.

Why does the air not experience dielectric breakdown in the case of the spike? How does current continue to flow through the air, without any sparks?

Answer 1

There are several types of discharge that might occur in a gaseous/plasma medium. The exact type of the discharge depends on several parameters: the type of gaseous medium (what kind of gas or plasma is present), pressure, voltage and current. Spark is what is called an arc discharge.

As far as I understand, in this case the type of medium and the pressure is the same in both experiments (ambient air at 1 atmosphere?). So the key difference was in voltage/current. The spike has lower capacitance with respect to the sphere, so it doesn't allow the voltage to build up enough to actually break down the air. In a somewhat simplified way one might look at it as such: the charge easily leaves the spike ("flows off" of it) because of it's point-sharp form whereas the sphere has plenty of room to accumulate the charge.

So as we increase voltage the current rises quicker in case of spike (yet it's not big enough to glow brightly), whereas the sphere accumulates a lot of charge with a relatively small current and then the air quickly breaks down with a spark. So, to sum up, very roughly speaking, in case of a spike we have relatively big current and low voltage, and in case of a sphere we have relatively low current and high voltage - which eventually leads to an arc.