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Van De Graff Current

by Armada420
Tags: current, graff
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Oct29-07, 01:29 AM
P: 1
1. The problem statement, all variables and given/known data
In a Van de Graaff electrostatic generator, a rubberized belt 30 cm wide travels at a velocity of 20 meters/sec. The belt is given a surface charge at the lower roller, the surface charge density being high enough to cause a field of 40 statvolts/cm on each side of the belt. What is the current in milliamps?

2. Relevant equations
I = nqua
E= 4(pi)kq?

3. The attempt at a solution
There might be some sort of Gaussian surface to be used here, I have the avg velocity (u), the area (a), the charge (q), just need the electrons per cubic centimeter to create such a field.
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Oct29-07, 06:32 AM
Astronuc's Avatar
P: 21,887
the surface charge density being high enough to cause a field of 40 statvolts/cm on each side of the belt
what is the surface charge [itex]\sigma[/itex] necessary to produce an electric field of 40 statvolts/cm on each side of the belt? The charge is carried on the surface of the belt, not throughout.
Oct29-07, 09:46 AM
HW Helper
P: 1,457
Electric charge is generated on the inside of belt at the bottom by contact with the roller (made of a different material) as it is driven over it. The field of this charge on the inside of the belt causes the brushes (sharp points with charge induced on them due to the presence of the field) next to the roller at the bottom to ionize the air (tips are charged oppositely to that of roller). Ions/charge out of the air around the brushes are then attracted towards the roller and sits on the outside of the belt. These are of opposite charge than that of the roller. It is these charges that are "removed" at the top of the belt with a similar mechanism (ionizing the air near the brushes at the top and being cancelled out). While the air is ionized at the top charges are fed into the brushes from the air, continually being conducted away to the sphere.

The belt therefore carries charges of opposite charge on its outsides surfaces and the situation looks very much like a capacitor with an electric field from the one to the other side.

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