if we leave this vandegraff running and don't discharge ..then will it keep on increasing charge and voltage? and if the dome of vandegraff is kept seperately in a vacuum than it would also keep on building the charge and voltage without auto discharge through dielectric breakdown of air forever?
The dielectric breakdown strength of air is 30,000 V/cm. Once the electric field at the surface of the dome gets to this electric field intensity, the voltage will not increase. You might not get a spark, but you will get corona discharge as the air ionizes and carries charge off the dome. I talk about the maximum voltage and show it is a function of the radius of the dome in this video th-cam.com/video/yaeITv9Ytko/w-d-xo.html
So, if you charge the leyden jar from the positive terminal of the V.G., it's positive that's stored in the jar, and negative that collects on the outside terminal? So if you hold the leyden jar, do you start accumulating negative charge?
Does that make one surrounded by a negative field when they hold the negative terminal? And/or, will you start attrracting negative ions by holding it?@@electricandmagneticfields2314
Is the amount of negative energy gathered upon a person proportional to the amount of positive juice in the jar? If I were to walk on a mountain or by a shore while holding the negative terminal - surely I would acquire a great deal of negative electrons, no? Would they just keep gathering until breakdown voltage is reached?
@@AlexanderStone I don't understand what you are asking. If you hold the jar, you are essentially acting as ground for the outer conductor of the capacitor. So if you add positive charge to the inner conductor, such as touching the ball to a positive voltage, electrons will flow from you to the outer wall of the conductor.
Sir, I have a question for you. Air has its own resistance and assuming that (I=V/R) is low enough not to kill you or make your heart skip a beat. I've also seen your Leyden jar video and the output of that cap had a current that can hurt you because (Q=V*C) and (I=dQ/dt). So, the bigger the Q, the bigger the I. If Q is bigger, then (Q=n*e) which means that n gets bigger too, right? Where n is the number of electrons. Would I be right in saying that the bigger I build a Van De Graaf generator or a Wimshurst machine, the greater the current? Or, the more of the little Van De Graaf generators or little Wimshurst machines I build, the more of the current is discharged? If I connected them in parallel to a main Leyden jar. Or do I need to connect them in series to add their voltages to increase the current because (I=V/R) and if V is bigger, then so is I and the resistance of air is constant, right?
That would be a good demo. I'll have to think about how to set that up, which probably would not happen till this summer. In the meantime, here is where I do talk about the photoelectric effect, th-cam.com/video/_kOZpu6OJs0/w-d-xo.html
Very great video, makes a lot of sense to me! I learned this in the very beginning of this semester but had no idea how it would be like in the actual world. Wow, physics is so true! I love it more! Luckily I’m majoring in physics!
Do you mean a second sphere that could be attached to ground and act as the second conductor of a capacitor with the sphere of the Van de Graaff generator being the other conductor? Then yes effectively you would have a much larger capacitance that would mean more charge for the same potential.
Hello, I have a question, If you would actually touch van de graaf, would it kill you? I made some calculations, lets say van der graaf has 300 000 V, assuming human body has a resistence of 2000, so the current is about 300 Amps
The current flows for a very short time as the sphere discharges. The buildup of charge on the sphere is small. Therefore if you are touching the sphere, the current flowing through your body is small.
@Electric and Magnetic Field. Sure, but isn’t it that current = charge per second. You assumed that the spark lasts 0.015 of second. I would say that electron flow is much more, about 200 000 000 meters/second, that makes the time over let’s say 0,1 m something like 5e-10 second. The current then will be something like 10^-6 charge / 5e-10 seconds which equals huge current. Even when I touch the outlet with 110V for a split of second it would give me a shock. I am confused what is actually harmful for human body. Is it the amount of charge going to your body or the speed of charge ? That is not the same, because hypotheticaly if I have a small amount of charge that have high speed it would give me a high current even though the amount of charge transfered is small.
Edit: I have consulted with my physics teacher at the university and turns out I was right. The current through your body is huge, but here the main role plays the amount of charge. So both of them are dangerous, its either big constant current circa > 3,5 mA) or big amount of accumulated charge that you instantly touch (circa > 50 microcoulomb) that is dangerous
@@michalkolarik9014 Since the sparks occur when I am about 10 cm away from the van de Graaff generator, the voltage is about 300, 000 V. (Dielectric breakdown strength of air is 30kV/cm.)
The capacitance of the van de Graaff generator is about 10 pF.
W = Energy stored on a capacitor = ½[C(V^^2)] = 0.45 J
About 1/2 a Joule, which is about the work down lifting an apple 1/2 meter is delivered to me with each spark.
Your channel is a gold mine. Thank you for the brilliant explanation!
You are welcome and thank you for the nice compliment!
Love from Russia!
@@bodiapa5720 Back at you!
if we leave this vandegraff running and don't discharge ..then will it keep on increasing charge and voltage? and if the dome of vandegraff is kept seperately in a vacuum than it would also keep on building the charge and voltage without auto discharge through dielectric breakdown of air forever?
The dielectric breakdown strength of air is 30,000 V/cm. Once the electric field at the surface of the dome gets to this electric field intensity, the voltage will not increase. You might not get a spark, but you will get corona discharge as the air ionizes and carries charge off the dome. I talk about the maximum voltage and show it is a function of the radius of the dome in this video th-cam.com/video/yaeITv9Ytko/w-d-xo.html
What is the upper roller and lower roller material?
upper roller material is glass and the lower one appears to be steel (axle).
So, if you charge the leyden jar from the positive terminal of the V.G., it's positive that's stored in the jar, and negative that collects on the outside terminal?
So if you hold the leyden jar, do you start accumulating negative charge?
Correct
Does that make one surrounded by a negative field when they hold the negative terminal? And/or, will you start attrracting negative ions by holding it?@@electricandmagneticfields2314
Is the amount of negative energy gathered upon a person proportional to the amount of positive juice in the jar? If I were to walk on a mountain or by a shore while holding the negative terminal - surely I would acquire a great deal of negative electrons, no? Would they just keep gathering until breakdown voltage is reached?
Or would I start pushing away negative electrons - being seen as a giant one to them? @@electricandmagneticfields2314
@@AlexanderStone I don't understand what you are asking. If you hold the jar, you are essentially acting as ground for the outer conductor of the capacitor. So if you add positive charge to the inner conductor, such as touching the ball to a positive voltage, electrons will flow from you to the outer wall of the conductor.
Sir, I have a question for you. Air has its own resistance and assuming that (I=V/R) is low enough not to kill you or make your heart skip a beat. I've also seen your Leyden jar video and the output of that cap had a current that can hurt you because (Q=V*C) and (I=dQ/dt). So, the bigger the Q, the bigger the I. If Q is bigger, then (Q=n*e) which means that n gets bigger too, right? Where n is the number of electrons. Would I be right in saying that the bigger I build a Van De Graaf generator or a Wimshurst machine, the greater the current? Or, the more of the little Van De Graaf generators or little Wimshurst machines I build, the more of the current is discharged? If I connected them in parallel to a main Leyden jar. Or do I need to connect them in series to add their voltages to increase the current because (I=V/R) and if V is bigger, then so is I and the resistance of air is constant, right?
yes, if voltage is higher, then the current is higher in the case that the material's resistance remains constant.
Wow really nice :)
Thank you!
Hi sir this is an Indian show I here8 requesting to do experiment of Hertz’s observations on photoelectric effect: so pls
That would be a good demo. I'll have to think about how to set that up, which probably would not happen till this summer. In the meantime, here is where I do talk about the photoelectric effect, th-cam.com/video/_kOZpu6OJs0/w-d-xo.html
wov
van de graaf merely continuously outpaces the relaxation time of the triboelectric materials allowing for redepositing of charges in a useful manner.
Sir hi can you upload the video for the doping of the material please
I don't understand what you are asking? Doping of what material? Are you talking about semiconductors?
Very great video, makes a lot of sense to me! I learned this in the very beginning of this semester but had no idea how it would be like in the actual world. Wow, physics is so true! I love it more! Luckily I’m majoring in physics!
Glad you enjoyed it and thanks for the kind comments.
If we put a metal sphere seperated by a dielectric on top of van degraff generator's top..would it keep on building high potential and charge density?
Do you mean a second sphere that could be attached to ground and act as the second conductor of a capacitor with the sphere of the Van de Graaff generator being the other conductor? Then yes effectively you would have a much larger capacitance that would mean more charge for the same potential.
very good video. i love how you combined two, fundamental inventions together.
Thanks
10:40 when the camera unfocuses, is that because of the sudeen light exposition or because of electromagnetic interactions?
Probably, there is an electromagnetic pulse that is probably effecting the electronics in the camera.
@@electricandmagneticfields2314cool
great video 🔥
Thanks!
Bro charging his arm
Don't do this at home!
Hello, I have a question, If you would actually touch van de graaf, would it kill you? I made some calculations, lets say van der graaf has 300 000 V, assuming human body has a resistence of 2000, so the current is about 300 Amps
The current flows for a very short time as the sphere discharges. The buildup of charge on the sphere is small. Therefore if you are touching the sphere, the current flowing through your body is small.
@Electric and Magnetic Field. Sure, but isn’t it that current = charge per second. You assumed that the spark lasts 0.015 of second. I would say that electron flow is much more, about 200 000 000 meters/second, that makes the time over let’s say 0,1 m something like 5e-10 second. The current then will be something like 10^-6 charge / 5e-10 seconds which equals huge current. Even when I touch the outlet with 110V for a split of second it would give me a shock. I am confused what is actually harmful for human body. Is it the amount of charge going to your body or the speed of charge ? That is not the same, because hypotheticaly if I have a small amount of charge that have high speed it would give me a high current even though the amount of charge transfered is small.
Edit: I have consulted with my physics teacher at the university and turns out I was right. The current through your body is huge, but here the main role plays the amount of charge. So both of them are dangerous, its either big constant current circa > 3,5 mA) or big amount of accumulated charge that you instantly touch (circa > 50 microcoulomb) that is dangerous
@@michalkolarik9014 It is the energy delivered, so integral of VIdt, which is small.
@@michalkolarik9014 Since the sparks occur when I am about 10 cm away from the van de Graaff generator, the voltage is about 300, 000 V. (Dielectric breakdown strength of air is 30kV/cm.)
The capacitance of the van de Graaff generator is about 10 pF.
W = Energy stored on a capacitor = ½[C(V^^2)] = 0.45 J
About 1/2 a Joule, which is about the work down lifting an apple 1/2 meter is delivered to me with each spark.