Breakdown voltage and electric field

In summary, the conversation is discussing the relationship between the breakdown voltage and radius of a conducting, charged sphere with a breakdown field of E_{breakdown}. The question is how to reduce the breakdown voltage by half, and the logic presented suggests that doubling the radius would achieve this. However, another person suggests that halving the radius would be the correct approach. The conversation ends with a summary of the formula V = Er and the conclusion that to halve the breakdown voltage, the radius must be halved.
  • #1
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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!
 
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  • #2
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?
 

What is breakdown voltage?

Breakdown voltage refers to the minimum voltage required to cause a sudden and significant increase in the flow of current through an insulating material, also known as an electrical breakdown. This can cause damage to the material and potentially lead to a short circuit.

What factors affect breakdown voltage?

Several factors can affect breakdown voltage, including the type and thickness of the material, the temperature, the presence of impurities, and the shape and size of the electrodes used to apply the voltage.

What is the relationship between breakdown voltage and electric field?

Breakdown voltage and electric field are directly proportional. This means that as the electric field increases, the breakdown voltage also increases. This relationship is described by the Paschen's law, which states that the breakdown voltage is inversely proportional to the product of the gas pressure and the distance between the electrodes.

Why is breakdown voltage important?

Breakdown voltage is an important concept in electrical engineering as it determines the maximum voltage that can be applied to a material before it breaks down. It is also important in the design and safety of electrical systems, as exceeding the breakdown voltage can lead to equipment failure and potential hazards.

How is breakdown voltage measured?

The breakdown voltage is typically measured using a high-voltage source, such as a transformer or a generator, and a voltage measuring instrument, such as a voltmeter. The voltage is gradually increased until the material breaks down, and the corresponding voltage is recorded as the breakdown voltage.

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