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- Homework Statement
- Hello, I was revising topics on electricity and fields when I found a question which although I have tried to answer I am having a little difficulty with. I would be very grateful if anyone could peruse my workings so far and suggest possible improvements, especially in regard to 1.iii.

The question follows:

It is found that air will ionise and conduct electric current when in an electric field greater than or equal to 3.0 x 10^6 Vm^-1.

Question 1;

i. A Van der Graaf generator has a top sphere with a radius of 30 cm. Find the greatest charge it can hold?

ii. Calculate the highest potential that the generator could be raised to?

iii. Suggest how the generator could be redesigned in order to raise it to a higher potential?

Question 2;

If the potential difference between two electrodes is 230 V, find how close they can they be before sparks travel between them, assuming that the electric field between them is uniform?

- Relevant Equations
- V=kQ/r

E=kQ/r^2

E=V/d

1. i. I think that the potential on the surface will be the same as that of a point charge at the centre of the Van der Graaf sphere, which will be 30cm away (since this is the radius of the top sphere). Convert 30cm to m which is equal to 0.3 m.

Therefore, to find the charge it can hold one can rearrange the equation for the electric field strength of a point charge E=kQ/r^2 which becomes Q=Er^2/k

Q=3*10^6*0.3^2/8.99*10^9

Q=3.003337041*10^-5 C ~ 3.00 *10^-5 C to 3.s.f or 30 μC

I know that I have calculated the charge here but would this be the

ii. Since the formula for the potential of a point charge is V=kQ/r, one may employ this in conjunction with the value calculated previously for the charge at 3.00 μC:

V=8.99*10^9*3.00 *10^-5/0.3

V=899,000 ~ 900,000 V or 900kV

(I noticed if I used the exact value calculated for the charge of 3.003337041*10^-5 C then the potential is equal to 900kV exactly without the need for rounding.)

Similarly, this is a calculation of the potential but is it the

iii. I am really stumbled here, I have not learned about Van der Graaf generators specifically before so I cannot broadly comment upon their structure. I understand from my own research that the maximum potential difference of a Van der Graaf generator can be increased by using a variation of the Van de Graaff accelerator called a tandem accelerator; in which ions are able to gain kinetic energy twice, firstly as negative ions and then as positive ions in a tandem fashion. However, I do not think that it the solution this question is searching for. Rather, perhaps I am supposed to comment upon how the potential could be increased using my knowledge of electric potential in a radial field, specifically focusing on the equation V=kQ/r.

Would I perhaps evaluate that the electric potential is directly proportional to the charge stored but inversely proportional to the separation distance. Therefore, by increasing the separation distance or rather the size of the top sphere could the Van der Graaff generator be raised to a higher potential? Additionally, by altering the charge could this also raise the potential, spec

Question 2.

Dry air is stated to support a maximum electric field strength of 3*10^6 Vm^-1. Above this point the field creates enough ionisation that the medium becomes a conductor, capable of facilitating a discharge or spark which would reduce the field.

V=230 V

E air = 3*10^6 Vm^-1

Therefore, for a uniform electric field, like that between two electrodes, E=V/d which can be rearranged as d=V/E.

Thus, d=230/3*10^6

d=7.666667*10^-5 m ~ 7.67 *10^-5m to 3.s.f

I would be incredibly appreciative of any help or feedback

Therefore, to find the charge it can hold one can rearrange the equation for the electric field strength of a point charge E=kQ/r^2 which becomes Q=Er^2/k

Q=3*10^6*0.3^2/8.99*10^9

Q=3.003337041*10^-5 C ~ 3.00 *10^-5 C to 3.s.f or 30 μC

I know that I have calculated the charge here but would this be the

**greatest**charge it can hold?ii. Since the formula for the potential of a point charge is V=kQ/r, one may employ this in conjunction with the value calculated previously for the charge at 3.00 μC:

V=8.99*10^9*3.00 *10^-5/0.3

V=899,000 ~ 900,000 V or 900kV

(I noticed if I used the exact value calculated for the charge of 3.003337041*10^-5 C then the potential is equal to 900kV exactly without the need for rounding.)

Similarly, this is a calculation of the potential but is it the

**highest**potential the generator could be raised to?iii. I am really stumbled here, I have not learned about Van der Graaf generators specifically before so I cannot broadly comment upon their structure. I understand from my own research that the maximum potential difference of a Van der Graaf generator can be increased by using a variation of the Van de Graaff accelerator called a tandem accelerator; in which ions are able to gain kinetic energy twice, firstly as negative ions and then as positive ions in a tandem fashion. However, I do not think that it the solution this question is searching for. Rather, perhaps I am supposed to comment upon how the potential could be increased using my knowledge of electric potential in a radial field, specifically focusing on the equation V=kQ/r.

Would I perhaps evaluate that the electric potential is directly proportional to the charge stored but inversely proportional to the separation distance. Therefore, by increasing the separation distance or rather the size of the top sphere could the Van der Graaff generator be raised to a higher potential? Additionally, by altering the charge could this also raise the potential, spec

Question 2.

Dry air is stated to support a maximum electric field strength of 3*10^6 Vm^-1. Above this point the field creates enough ionisation that the medium becomes a conductor, capable of facilitating a discharge or spark which would reduce the field.

V=230 V

E air = 3*10^6 Vm^-1

Therefore, for a uniform electric field, like that between two electrodes, E=V/d which can be rearranged as d=V/E.

Thus, d=230/3*10^6

d=7.666667*10^-5 m ~ 7.67 *10^-5m to 3.s.f

I would be incredibly appreciative of any help or feedback