Calculate work done on charged object between plates

In summary, the voltage difference between the plates would be 3kV, but the electric field between the plates would be much higher, at around 230769.23 N/C.
  • #1
EM da Costa
2
0

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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  • #2
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?
 

1. How do you calculate the work done on a charged object between two plates?

The work done on a charged object between two plates can be calculated by multiplying the charge of the object by the potential difference between the plates. This is represented by the formula W = qΔV, where W is the work done, q is the charge of the object, and ΔV is the potential difference between the plates.

2. What units are used to measure work done on a charged object?

The units used to measure work done on a charged object are joules (J). This is the standard unit for measuring energy in the International System of Units (SI).

3. Can the work done on a charged object be negative?

Yes, the work done on a charged object can be negative. This occurs when the object is moving in the opposite direction of the electric field between the plates. This means that energy is being transferred from the object to the system, resulting in a decrease in the object's energy.

4. How does the spacing between the plates affect the work done on a charged object?

The spacing between the plates does not directly affect the work done on a charged object. However, it does affect the potential difference between the plates, which in turn affects the work done. As the spacing between the plates increases, the potential difference decreases, resulting in a decrease in the work done on the charged object.

5. Can the work done on a charged object between plates be used to determine the electric field strength?

Yes, the work done on a charged object between plates can be used to determine the electric field strength. This can be done by rearranging the formula for work done to solve for the electric field intensity, which is represented by the formula E = W/qd, where E is the electric field intensity, W is the work done, q is the charge of the object, and d is the distance between the plates.

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