# Homework Help: Potential Difference in a van der graaff generator

1. Oct 4, 2011

### JNBirDy

Hi, this problem is from 'Electromagnetism' by Grant & Phillips and it states,

1. The problem statement, all variables and given/known data
'http://i1129.photobucket.com/albums/m505/physicsbird1/126fig.png" [Broken] shows a cross-section of the cylindrical high-voltage terminal of a van der Graaff generator, surrounded by an 'intershield' and a pressure vessel, both of which are also cylindrical. The gas in the pressure vessel breaks down in electric fields greater than 1.6 x 10$^{7}$ volts/m. If the radii of the terminal, intershield and pressure vessel are 1.5 m, 2.5 m and 4 m respectively, what is the highest potential difference that can be maintained between the terminal and the pressure vessel?'

The answers given in the back of the book: '31.1 MV. When the terminal is at this voltage the intershield must be at 18.8MV to prevent breakdown.'

2. Relevant equations
d∅ = -E$\bullet$dr

E = $q/4πε₀r^{2}$

3. The attempt at a solution

1.6x10$^{7}$ = $q/4πε₀(2.5)^{2}$

1.00x10$^{8}$ = $q/4πε₀$ (Let $q/4πε₀$ = A)

A = 1.00x10$^{8}$

∅$_{B}$ - ∅$_{A}$ = -1.00x10$^{8}$$\int^{4}_{1.5}$(1$/r^{2})$

= -1.00x10[itex]^{8}[-1/r]^{4}_{1.5}

= -1.00x10^{8} [-1/4 + 1/1.5]

= -4.17x10^{7}

So... can anybody tell me where I've went wrong?

Last edited by a moderator: May 5, 2017
2. Oct 5, 2011

3. Oct 5, 2011

### Spinnor

You wrote, "E = q/4πε₀r^2"

E goes as 1/r ?