Parallel plate electric fields -- # of electrons transferred

In summary, an oil droplet is suspended between two horizontal parallel plates with a separation of 0.4 cm. A potential difference of 320V is applied to the plates. Using the equations FE = (k⋅q1⋅q2)/r^2, E = V/d, and V = (k⋅q)/r, it is possible to find the charge (q) of the oil droplet and divide it by the charge of one electron (1.602×10^-19 C) to determine the number of electrons transferred. However, considering the excessive amount of electrons calculated (887779164.9), this method may not be accurate. Further investigation and analysis may be needed.
  • #1
radiant_june
2
1

Homework Statement


An oil droplet is suspended between two horizontal parallel plates with a separation of 0.4 cm. If the potential difference of 320V is applied to the plates, determine the number of electrons transferred to/from the droplet.

Given/Known Values
mdroplet = 5.2×10-6 kg
d = 0.4 cm = 0.004 m (distance between plates)
V = 320 V
k = 9.0×109 Nm2/C2

Homework Equations


Equations
FE = (k⋅q1⋅q2)/r2
E = V/d
E = FE/qtest
V = (k⋅q)/r
1 electron = 1.602×10-19 C

The Attempt at a Solution


My assumption was that I could find the charge (q) of the oil droplet, using the V = (k⋅q)/r formula. Then I could divide q by 1.602×10-19 to get the number of electrons transferred.

V = (k⋅q)/r
320 = (9.0×109⋅q)/0.004
1.28/9.0×109 = q
q = 1.422×10-10

electrons = q/1.602×10-19
= 1.422×10-10/1.602×10-19
= 887779164.9 electrons

Considering the excessive amount of electrons, I do not think I did this correctly. Any help would be appreciated!

 
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  • #2
The given voltage V is the potential difference between the two metal plates. This establishes an electric field between the plates. What forces are operating on the oil drop? Can you draw a Free Body Diagram for it?
 
  • #3
As worded, looks like a trick question to me ...
 

1. What is a parallel plate electric field?

A parallel plate electric field is a type of electric field that exists between two parallel conducting plates. It is created when a potential difference is applied between the two plates, causing an electric field to form between them.

2. How does a parallel plate electric field transfer electrons?

In a parallel plate electric field, electrons are transferred from one plate to the other due to the difference in potential between the two plates. Electrons are negatively charged and will flow from the plate with a higher potential to the plate with a lower potential.

3. What factors affect the number of electrons transferred in a parallel plate electric field?

The number of electrons transferred in a parallel plate electric field is affected by the amount of potential difference between the plates, the distance between the plates, and the material of the plates. A higher potential difference and smaller distance between the plates will result in a larger number of electrons being transferred. Additionally, the type of material used for the plates can affect the transfer of electrons.

4. How can the number of electrons transferred in a parallel plate electric field be calculated?

The number of electrons transferred in a parallel plate electric field can be calculated using the formula Q = C*V, where Q is the charge (in coulombs), C is the capacitance (in farads), and V is the potential difference (in volts) between the plates. The capacitance can be calculated using the formula C = ε*A/d, where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates.

5. Can a parallel plate electric field be used for practical applications?

Yes, parallel plate electric fields have many practical applications, such as in capacitors, particle accelerators, and electronic devices. They can also be used for air purification, electrostatic painting, and electrostatic discharge protection.

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