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Breakdown voltage and electric field

  • #1
798
1

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 [itex]E_{breakdown}[/itex], 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:

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

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

Thanks in advance!
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
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?
 

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