Question about Van De Graaf generator

  • Thread starter Thread starter Parad0x88
  • Start date Start date
  • Tags Tags
    Generator
AI Thread Summary
The discussion revolves around understanding the maximum potential of a metallic sphere on a Van de Graaff generator. The textbook states that the breakdown electric field in air limits a sphere of 1.00 m radius to a maximum potential of 3 X 10^6 V. For a sphere with a radius of 2.0 m carrying a charge of 3 x 10^-5 C, there is confusion about whether to simply double the potential or apply the equation V = Q/4∏rε0 to calculate the voltage. Clarification is provided that the equation is necessary to determine the actual voltage, as the text's line indicates the maximum potential before breakdown occurs. Understanding this distinction is crucial for safe operation of the generator.
Parad0x88
Messages
73
Reaction score
0

Homework Statement


I have a sentence in my textbook that reads:

Because the "breakdown" electric field in air is about 3 X 106 V/m, a sphere 1.00m in radius can be raised to a maximum potential of 3 X 106 V

And now the problem states: The metallic sphere on top of a large van de Graaf generator has a radius of 2.0 m. Suppose that the sphere carries a charge of 3x10-5 C

Homework Equations


V = Q/4∏rε0

The Attempt at a Solution


Does that mean to find the Volt, I have to simply multiply that 3 X 106 V by 2, to give me 6 X 106 V?

Or do I have to use the above equation to find the volt inside the generator?
 
Physics news on Phys.org
Why not employ your Relevant Equation?
 
gneill said:
Why not employ your Relevant Equation?

I guess it's the line from the text that confused me as to whether that equation was even necessary or not
 
The line in the text is telling you what limits the maximum potential on a surface (before breakdown and arcing occur to drain off the charge). If your calculated potential approaches this value, stand well back from the apparatus!:eek:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top