Electric charge to cause spark in air

In summary: Paths of ionised air will form between the spheres if the electric field is strong enough.But it's not a requirement that the whole 'path' from one sphere to the other needs to have the field, only a small region of it.In summary, when two charged metal spheres are brought together (in a Wimshurst machine), a spark occurs at a distance of 1.5cm. An electric field of 30,000V per cm is required to cause a spark.
  • #1
flash
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When two charged metal spheres are brought together (in a Wimshurst machine), a spark occurs at a distance of 1.5cm. An electric field of 30,000V per cm is required to cause a spark. I need to work out the charge on the two balls, assuming (I think) it is equal on both.

I know how to calculate the field at any point from each sphere, but not sure how to work backwards and find the charge.
 
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  • #2
Well, since the spheres are conducting the charge is distributed uniformally on the surface of the sphere, so we can treat the two spheres as point charges. Tell me, the distance you quote is that measured from the centre of the two spheres?
 
  • #3
Yeah I planned to treat them as point charges, but the distance quoted between the two surfaces. The spheres are 2cm in diameter. I don't know how to calculate the charges even if they were point charges :(
 
  • #4
Okay, so if we consider the spheres as to point charges, what is the distance between the two point charges?

Also, what is the definition of the electric field?
 
  • #5
The distance between the centre of the spheres is 3.5cm.
Electric field is an area where a charge experiences a force (F=Eq). The field strength from a point is kq/r^2 so the field strength between the points would be the vector sum of the field from each sphere. If they are equal and opposite charges this is just 2kq/r^2. I don't think the two spheres will create a uniform field though, which the question sort of implies.
 
  • #6
flash said:
The distance between the centre of the spheres is 3.5cm.
Electric field is an area where a charge experiences a force (F=Eq). The field strength from a point is kq/r^2 so the field strength between the points would be the vector sum of the field from each sphere. If they are equal and opposite charges this is just 2kq/r^2. I don't think the two spheres will create a uniform field though, which the question sort of implies.
All is correct. However, you can still calculate the potential of some point in an elctric field can you not?
 
  • #7
The potential difference V is the product of the field strength and the distance (I could be wrong), V=Ed. The question says the air ionises when the field strength is 30,000V/cm which means E=(30000/0.015)=2*10^6 N/C. So the air ionises causing a spark when there is a field of that magnitude between the two spheres. But if the field is not uniform, that doesn't make sense.
 
  • #8
flash said:
The potential difference V is the product of the field strength and the distance (I could be wrong), V=Ed. The question says the air ionises when the field strength is 30,000V/cm which means E=(30000/0.015)=2*10^6 N/C. So the air ionises causing a spark when there is a field of that magnitude between the two spheres.
Correct!
But if the field is not uniform, that doesn't make sense.
Why must the field be uniform? What would you see if the field was uniform?
 
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  • #9
If the field isn't uniform then some parts of the air are ionised and other arent. Does the spark only occur when the minimum field strength (I guess its in the middle of the spheres?) is 2*10^6 N/C? I was assuming that it would be uniform so I could say, when all the area between the spheres had a field of x, the air ionised and a spark happened.
 
  • #10
flash said:
If the field isn't uniform then some parts of the air are ionised and other arent.
Correct :approve:
I was assuming that it would be uniform so I could say, when all the area between the spheres had a field of x, the air ionised and a spark happened.
You can still say this, except that it is not a requirement that the whole area between the spheres have the required electric field, only a small region of it.
 
  • #11
Does the whole 'path' from one sphere to the other need to have this field? I just made a graph of position between the spheres vs field strength and is looks the shape of the letter u. The minimum field is in the centre of the two spheres so if a path of 'ionised air' is to be made, this would be the area last to be ionised and hence what I use to calculate the answer.
 
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FAQ: Electric charge to cause spark in air

1. What is electric charge?

Electric charge is a fundamental property of matter that results from the presence of charged particles, such as protons and electrons. It is measured in Coulombs (C) and can be either positive or negative.

2. How does electric charge cause a spark in air?

When an object or surface has a high concentration of electric charge, it can create a strong electric field. If this electric field is strong enough, it can ionize the air molecules around it, causing them to lose or gain electrons. This creates a conductive path for the electric charge to travel through, resulting in a spark.

3. What determines the strength of the spark?

The strength of the spark is determined by the amount of electric charge present and the strength of the electric field. The greater the electric charge and the stronger the electric field, the more powerful the spark will be.

4. Can any type of electric charge cause a spark in air?

No, only high concentrations of electric charge can cause a spark in air. This is because the electric field needs to be strong enough to ionize the air molecules. Small amounts of electric charge may not produce enough energy to overcome the resistance of the air and create a spark.

5. Is a spark in air dangerous?

A spark in air can be dangerous if it occurs in an environment where there are flammable materials present. This is because the heat from the spark can ignite the flammable material and cause a fire or explosion. It is important to always handle and control electric charges carefully to avoid any potential hazards.

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