Breakdown voltage and electric field

[sandy.bridge]

Homework Statement

Hey guys. I am looking for confirmation regarding my logic concerning a specific principle. If I have a conducting, charged sphere that has a breakdown field of $E_{breakdown}$, and I want to know what the radius of the sphere has to be set to in order to reduce the breakdown voltage by half, would the following logic be correct:

$V_{breakdown} ∝ 1/r$ and therefore $(1/2)V_{breakdown} ∝ 1/(2r)$. Hence to half the breakdown voltage, the radius is doubled.

$dV=\vec{E}.\vec{dl}$, where for the specific radius $2r$, dl ends up at the radius, hence halving the potential means that the field is also halved. Hence,
$\vec{E}_{breakdown}/(2r)^2=0.5\vec{E}_{breakdown}$, and $r=0.7071$.

Thanks in advance!

 

[gneill]

Hmm. I'd have thought that doubling the radius would double the breakdown voltage so that $V_b ∝ r$ ; Bigger spheres allow greater potential before sparks fly.

Here's my thoughts:

For a spherical charged surface:

$E = k \frac{Q}{r^2}~~~~$ and $~~~~ V = k \frac{Q}{r}$

yieding V = Er

Assuming that it's the field strength that's responsible for breakdown, for a given breakdown field strength $E_b$, the corresponding breakdown voltage is

$V_b = E_b r$

So to halve the breakdown voltage, halve the radius. Does that make sense?