Calculate work done on charged object between plates

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SUMMARY

The discussion focuses on calculating the theoretical time for a piece of aluminum foil to travel between two charged plates with a potential difference of 3kV, created by a Cockcroft–Walton generator. The user applied key equations, including W = F.d and V = E.d, to derive the electric field strength of 230769.23 N/C and charge density of 230905.35 C/m². However, discrepancies arose when calculating the force between the plates, leading to an unrealistic charge of ±979.27C. The user seeks clarification on their assumptions regarding voltage constancy and equation application.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with the Cockcroft–Walton generator
  • Knowledge of basic physics equations related to work and force
  • Concept of charge density and permittivity of air
NEXT STEPS
  • Review the principles of electric fields and their calculations
  • Explore the effects of drag on charged objects in an electric field
  • Investigate the Cockcroft–Walton generator's operation and applications
  • Learn about charge distribution and its impact on electric forces
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators designing experiments to demonstrate electric field concepts.

EM da Costa
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Homework Statement


I am trying to calculate the theoretical time that it takes for a small piece of aluminium foil on a bottom plate to reach a top plate. The plates have a potential difference created by a 3kV Cockcroft–Walton generator. The plates are short cylinders with a surface area of 0.004241m2 and they are 0.013m apart from each other. The piece of aluminium foil is 1x0.5x0.0016cm(height, length, thickness) and it's density is 2.7g/cm3. I am going to compare the result of this calculation with the real experimental result. This is for a school assignment, the teacher asked us to design an experiment and to compare the two results to 'prove' the equations we're studying so i decided to use my voltage multiplier for the experiment.

Homework Equations


W = F.d
F = q.E
V = E.d
V = K.Q/d
W = q.U
Air permittivity = 1.00058986
E = charge density / air permittivity
charge density = Q/A

The Attempt at a Solution


So i know that the key is to determine the work done. So first i started with trying to calculate the electric field between the two plates. Of course i suppose that in this circuit the differential voltage of the two plates is not going to change with distance, charge is, so differential voltage is assumed to be of 3000v. On that assumption i calculated the electric field with the equation V = E.d to be 230769.23 N/C. After that i applied the equation E = charge density / air permittivity to calculate the charge density of the plate. My result was 230905.35 C/m2 and combining it with the surface area of the plate (0.004241m) i calculated that one of the plates were charged with 979,27C and the other one with -979,27C. That's when i knew i did something wrong, i used the charges of the plates to calculate the force between then and i got a result with a very high force that was very far from the force i felt while doing the experiment. I don't know if i was wrong with the assumption that the voltage doesn't change or if i combined one of the equations poorly. Help me please.
 
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I see this thread is marked as solved. Is that in error?
EM da Costa said:
the key is to determine the work done.
In a vacuum, yes, but a thin piece of foil is going to lose a lot to drag.
EM da Costa said:
230769.23
That seems about ten times too much.
EM da Costa said:
and the other one with -979,27C.
I assume that is a decimal comma. But are you not doubling up here? You calculated the voltage difference. If you are going to say that one is at voltage +V and the other at -V, what does that make V?
 

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