Electric field and charge density on a Van de Graaff dome with Gauss' law

[flash]

Homework Statement

Use Gauss' law to find the charge density on a Van Der Graff dome (r=40cm) if it is charged to 100kV. What is the electric field strength at r=25cm?

I understand gauss's law and I know that I need to use it to find the total charge enclosed on the dome. I can then work out the surface area of the dome and calculate charge density.

If I define a gaussian surface to be a sphere outside the dome, I can work out the charge enclosed if i know the electric field at the gaussian surface. I figure I must use the 100kV given to find the field at the gaussian surface, but I don't understand how.

Also, I need a little help with the second part of the problem. That radius given is inside the dome, unsure of where to go there.

Thanks for any help.

 

[Mindscrape]

So, as a starting point, what are the equations that you have to work with?

Also, as a matter of curiosity, what level of physics is this (I'm assuming a university calc based intro E&M), and what math have you taken?

 

[jackiefrost]

Check out http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potsph.html"

 

[nrqed]

Homework Statement

Use Gauss' law to find the charge density on a Van Der Graff dome (r=40cm) if it is charged to 100kV. What is the electric field strength at r=25cm?

I understand gauss's law and I know that I need to use it to find the total charge enclosed on the dome. I can then work out the surface area of the dome and calculate charge density.

If I define a gaussian surface to be a sphere outside the dome, I can work out the charge enclosed if i know the electric field at the gaussian surface. I figure I must use the 100kV given to find the field at the gaussian surface, but I don't understand how.

Also, I need a little help with the second part of the problem. That radius given is inside the dome, unsure of where to go there.

Thanks for any help.

That's probably a stupid question but I am wondering...is the sphere meant to be hollow or full? If it's hollow (which makes more sense), then the E field inside the dome will be zero since there is no net charge enclosed (I am assuming a complete sphere, in an actual VdG there is of course a part of the sphere cut out but here I assume they mean to ignore that). The 100 kV is not needed. So I must not understand the question!

 

[siddharth]

That's probably a stupid question but I am wondering...is the sphere meant to be hollow or full? If it's hollow (which makes more sense), then the E field inside the dome will be zero since there is no net charge enclosed (I am assuming a complete sphere, in an actual VdG there is of course a part of the sphere cut out but here I assume they mean to ignore that). The 100 kV is not needed. So I must not understand the question!

I think Van der Graaff generators usually use a hollow sphere. In any case, they use a metal sphere, so any charge would be on the outside of the sphere.

 

[nrqed]

I think Van der Graaff generators usually use a hollow sphere. In any case, they use a metal sphere, so any charge would be on the outside of the sphere.

Ok. So whether it is a hollow sphere or a full conducting sphere, the E field at 25 cm would be zero with no calculation required. (I guess the 100 kV was just needed to find the total charge)

Thanks for your input

 

[jackiefrost]

... the E field at 25 cm would be zero with no calculation required. (I guess the 100 kV was just needed to find the total charge)

Isn't it interesting how the E field is zero on the inside of a [perfectly enclosed] hollow conductive sphere? This why the VanDeGraff was designed to deposit charge from the inside rather than outside. Imagine how difficult it would be to deliver charges from the outside, always working against the field produced by the previously deposited charges.