Spherical balloon with conductive coating

[meteorologist1]

Hi, I have no idea of how to do the following problem and what formulas I should use. Please help! Thank you.

A spherical balloon has a conductive coating and we propose to inflate the balloon to a diameter of 0.1 meters by connecting the surface to a high voltage source. Suppose that the maximum practical electric field at the surface is 2 x 10^6 volts/meter (in air, just before breakdown of air molecules). What is the largest voltage we can apply, and what is the outward pressure (N/meters^2)? How many atmospheres is this?

 

[learningphysics]

Hi, I have no idea of how to do the following problem and what formulas I should use. Please help! Thank you.

A spherical balloon has a conductive coating and we propose to inflate the balloon to a diameter of 0.1 meters by connecting the surface to a high voltage source. Suppose that the maximum practical electric field at the surface is 2 x 10^6 volts/meter (in air, just before breakdown of air molecules). What is the largest voltage we can apply, and what is the outward pressure (N/meters^2)? How many atmospheres is this?

Find the total charge Q on the sphere in terms of the electric field - gauss' law works here.

Then find the voltage on the surface of the sphere in terms of Q... plug in Q from the first part, to get the voltage.

 

[meteorologist1]

Ok I understand the first part -- I can find the charge from Gauss's Law. For the second part, I'm still not sure what equation I should use. Should I use the equation W=QV or some other work equation? What about the outward pressure?

 

[meteorologist1]

Ok when I try it: using 1) E = kq/r^2 and 2) V = kq/r, I get V = Er = (2 x 10^6 volts/m)(0.1 m) = 2 x 10^-7 volts. This procedure looks too simple. Is it right?

 

[learningphysics]

Ok when I try it: using 1) E = kq/r^2 and 2) V = kq/r, I get V = Er = (2 x 10^6 volts/m)(0.1 m) = 2 x 10^-7 volts. This procedure looks too simple. Is it right?

Yes, those two formulas work because of the symmetry involved. But r=0.05m, so V=Er=2*10^6*0.05=100,000V.

I'm not sure about the pressure part... if Force on a small area dA is F=(sigma*dA)E (where sigma is charge density)... And then you can find pressure by F/dA.

But using F=(sigma*dA)E seems wrong to me as the charges are located inside the conductor where electric field is zero... Not sure here. Sorry!