Van der Graaf Generator: Charge, Potential and Redesign Questions

In summary, the conversation discusses the calculation of the charge and potential of a Van der Graaf generator, as well as the maximum electric field strength that can be supported by dry air. The potential on the surface of the generator can be found by rearranging the equation for electric field strength of a point charge and is equal to 900kV. However, it is not the highest potential the generator can be raised to. The electric field strength above the charged sphere is largest at the surface and the corresponding potential can be found using V=kQ/r. The maximum electric field strength supported by dry air is 3*10^6 Vm^-1, resulting in a distance between two electrodes of ~0.1mm.
  • #1
AN630078
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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 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 👍
 
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  • #2
First you need to learn LATEX and use it. It is an ongoing process.
Second please do not report numbers to 10 significant figures for no reason.
Now to the specifics
1) You need to know the field outside the charged sphere. How do you find the field of a charged sphere?? It will be largest right at the surface
2)What is the corresponding potential ?
3)this should be obvious from (1) and (2). This is a sphere and so things are easier.

Part 2 seems reasonable to me. And ~0.1 mm seems a good answer
 
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  • #3
hutchphd said:
First you need to learn LATEX and use it. It is an ongoing process.
Second please do not report numbers to 10 significant figures for no reason.
Now to the specifics
1) You need to know the field outside the charged sphere. How do you find the field of a charged sphere?? It will be largest right at the surface
2)What is the corresponding potential ?
3)this should be obvious from (1) and (2). This is a sphere and so things are easier.

Part 2 seems reasonable to me. And ~0.1 mm seems a good answer
Thank you for your very swift reply 😁. Yes confessedly I do need to learn LATEX! So would I assume my solutions for question 1 are incorrect?
The reason I quoted to such a extensive length of siginifcant figures was to show that the solutions I obtained convereged to whole solutions that did not require rounding (e.g. the potential exactly corresponded to 900kV if the charge is quoted to an appreciable length of significant figures but had to be rounded to 900kV otherwise).
The electric field strength of a point charge can be found using E=kQ/r^2, is this what you are referring to?
Would the corresponding potential be found using V=kQ/r?
 
  • #4
This is not a point charge. I believe you understand but you need to say the "magic words". I think your answer incorrect and I know it is incomprehensible.
 
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  • #5
hutchphd said:
This is not a point charge. I believe you understand but you need to say the "magic words". I think your answer incorrect and I know it is incomprehensible.
Thank you very much for taking the time to reply. Haha, yes I would have to agree, perhaps I have misunderstood the question and am searching for the "magic words". I am a little at a standstill though do you have any suggestions? 😊
 
  • #6
hutchphd said:
You need to know the field outside the charged sphere.
@AN630078 wrote that it will be the same as for that distance from an equal point charge, which is true. The 30μC looks right to me.
AN630078 said:
would this be the greatest charge it can hold?
If the charge were greater, what would happen?
AN630078 said:
calculated previously for the charge at 3.00 μC:
Typo: 30.0 not 3.00
I confirm 900kV. You could get there by combining the formulae to produce V=rE etc.
AN630078 said:
by increasing ... the size of the top sphere could the Van der Graaff generator be raised to a higher potential?
What do you think?

Your answer to Q 2 is good too.
 
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  • #7
haruspex said:
@AN630078 wrote that it will be the same as for that distance from an equal point charge, which is true. The 30μC looks right to me.

If the charge were greater, what would happen?

Typo: 30.0 not 3.00
I confirm 900kV. You could get there by combining the formulae to produce V=rE etc.

What do you think?

Your answer to Q 2 is good too.
Thank you very much for your reply also.
If the charge were greater would the generator begin to produce sparks?
Thank you, yes sorry I missed that typo, and thank you for suggesting combining the formula to produce V=rE.
I am still not sure about the redesign of the generator to raise the potential. My earlier suggestions were merely thoughts I gathered after considering the formula for the potential.
Would high voltages be limited by charge leakage currents on account of the conductivity of the air, which would increase as a result of the air molecules being ionised within high electric fields at the surface of the sphere causing spark discharges?

Thank you for evaluating question 2.

I am a little confused from the feedback I have received though, would my answer to question 1 be utterly incorrect since the sphere of the generator is not a point charge as stated by @hutchphd?
 
  • #8
AN630078 said:
I am still not sure about the redesign of the generator to raise the potential
You asked specifically whether
AN630078 said:
increasing ... the size of the top sphere could the Van der Graaff generator be raised to a higher potential?
I am asking you whether you think it would. What do the equations tell you?
AN630078 said:
would my answer to question 1 be utterly incorrect since the sphere of the generator is not a point charge
My guess is that @hutchphd missed, misinterpreted or disagreed with this statement in post #1:
"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)."
I believe that is correct.
 
  • #9
My original concern with the OP response for the electric field was that he nowhere justified the use of the point charge E field by saying the magic words "Using Gauss's Theorem" we know the E field. Or that it is the gradient of the potential. I think the correct number was reached (as you ascertained) but frankly I was (and remain) unwilling to plow through calculations that require me to work it out. I call it constructive laziness. Let the OP show us the correct answer clearly if indeed it exists.
That being said I was a trifle harsh. I did say "I think" the answer was incorrect...I meant I couldn't follow his solution and therefore inferred it likely incorrect. My apologies.
But learn LATEX. A little at a time.
 
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Related to Van der Graaf Generator: Charge, Potential and Redesign Questions

1. What is a Van der Graaf Generator?

A Van der Graaf Generator is an electrostatic generator that uses a moving belt to accumulate and transfer electric charge to a large metal sphere. The sphere becomes highly charged and can create large electric potentials.

2. How does a Van der Graaf Generator work?

A Van der Graaf Generator works by using a moving belt to transfer electric charge from a high voltage source to a large metal sphere. The belt is made of a non-conductive material and is constantly moving, creating friction and transferring charge to the sphere. This creates a high electric potential on the sphere, which can then be used for experiments or demonstrations.

3. What is the purpose of a Van der Graaf Generator?

A Van der Graaf Generator is primarily used for scientific experiments and demonstrations involving high electric potentials. It can also be used for particle acceleration and radiation therapy in medical settings.

4. Can a Van der Graaf Generator be redesigned to generate even higher potentials?

Yes, a Van der Graaf Generator can be redesigned to generate even higher potentials by increasing the size of the sphere and the speed of the belt. However, there are limits to how high the potential can be, as it is dependent on factors such as the material of the sphere and the surrounding environment.

5. Is a Van der Graaf Generator safe to use?

When used properly and with proper safety precautions, a Van der Graaf Generator is generally considered safe to use. However, it can generate very high electric potentials and should be handled with caution. It is important to follow all safety guidelines and instructions when operating a Van der Graaf Generator.

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