Electric charge to cause spark in air

[flash]

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.

 

[Hootenanny]

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?

 

[flash]

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 :(

 

[Hootenanny]

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?

 

[flash]

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.

 

[Hootenanny]

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?

 

[flash]

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.

 

[Hootenanny]

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?

 

[flash]

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.

 

[Hootenanny]

If the field isn't uniform then some parts of the air are ionised and other arent.

Correct

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.

 

[flash]

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.