Question about Van De Graaf generator

  • Thread starter Thread starter Parad0x88
  • Start date Start date
  • Tags Tags
    Generator
Click For Summary

Homework Help Overview

The discussion revolves around the potential of a metallic sphere on a Van de Graaff generator, particularly focusing on the relationship between charge, radius, and electric potential as described in a textbook. The problem involves understanding the implications of the breakdown electric field in air and how it relates to the maximum potential that can be achieved by the sphere.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore whether to simply double the maximum potential stated in the textbook or to apply a specific equation involving charge and radius to calculate the potential. There is confusion regarding the necessity of the equation based on the textbook's statement.

Discussion Status

Some participants have suggested using the relevant equation to clarify the situation, while others have pointed out the significance of the textbook's statement regarding maximum potential and its implications. The discussion reflects a mix of interpretations and considerations without reaching a consensus.

Contextual Notes

Participants are navigating the constraints of the problem, including the maximum potential before breakdown occurs and the specific charge and radius of the sphere in question. There is an acknowledgment of the potential risks associated with high voltages as indicated in the discussion.

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:
 

Similar threads

Replies
8
Views
3K
  • · Replies 12 ·
Replies
12
Views
7K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
1K