How do I calculate the acceleration of a particle in a parallel plate capacitor?

In summary, the conversation discusses finding the acceleration of a particle after a potential is applied, using the equation v_f^2 = v_i^2 + 2ad. It is determined that this equation only applies for constant acceleration, and the correct approach is to use qE = ma. The electric field between the plates is found using the net charge and density of the droplet, and is equal to twice the force of gravity. An expression for the magnitude of the electric field is given as E = \frac{2mg}{q}.
  • #1
temaire
279
0

Homework Statement


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The Attempt at a Solution


All I need to know is the acceleration acting on the particle after the potential is applied.

[itex]v_f^2 = v_i^2 + 2ad[/itex]
[itex](1*10^{-5})^2 = -(1*10^{-5})^2 + 2a(1*10^{-3})[/itex]
[itex]a = 1*10^{-7}[/itex] m/s2

Is this the correct approach to finding acceleration?
 
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  • #2
I don't think so because your acceleration is actually not going to me constant. Recall that the equation you used only applies for a constant acceleration.

I think you should go along the lines of qE = ma to find this.

You're given enough info to find the electric field between the plates, you are given the net charge of the droplet, and density multiplied by volume will give you the mass of your droplet.
 
  • #3
How do I find the electric field between the plates? I don't have the charge of the plates.
 
  • #4
Draw a free body diagram for the oil droplet for both of the given cases. What forces are acting? Can you draw any conclusions about the magnitude of the force due to air friction when the droplet is moving at speed vt?
 
  • #5
By drawing a FBD for the first case, I was able to determine the force due to air friction is equal to force due to gravity. For the second FBD, I was able to determine that the force due to the electric field is equal to the sum of the forces due to air friction and gravity. Therefore, the force due to the electric field is equal to twice the force due to gravity. Is this correct?
 
  • #6
temaire said:
By drawing a FBD for the first case, I was able to determine the force due to air friction is equal to force due to gravity. For the second FBD, I was able to determine that the force due to the electric field is equal to the sum of the forces due to air friction and gravity. Therefore, the force due to the electric field is equal to twice the force due to gravity. Is this correct?

Sounds good!

What does that tell you about the magnitude of the electric field between the plates (when the plates are given a potential difference)? Can you write an expression?
 
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  • #7
[itex]E = \frac{2mg}{q}[/itex]

Thanks for the help.
 

Related to How do I calculate the acceleration of a particle in a parallel plate capacitor?

1. What is a parallel plate capacitor?

A parallel plate capacitor is a device used to store electrical energy. It consists of two parallel conductive plates separated by a dielectric material. When a voltage is applied to the plates, an electric field is created between them, resulting in the storage of electrical charge.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by utilizing the principle of capacitance, which is the ability of a system to store electrical charge. When a voltage is applied to the plates, the electric field between them causes electrons to accumulate on one plate, while the other plate becomes positively charged. This results in the storage of electrical energy between the plates.

3. What is the equation for calculating the capacitance of a parallel plate capacitor?

The equation for calculating the capacitance of a parallel plate capacitor is C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material between the plates, A is the area of the plates, and d is the distance between the plates.

4. What factors can affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be affected by several factors, including the distance between the plates, the area of the plates, and the dielectric material used. The type of material used for the plates and the voltage applied can also have an impact on the capacitance.

5. What are some real-life applications of parallel plate capacitors?

Parallel plate capacitors have many real-life applications, including in electronic devices such as computers and smartphones. They are also used in power systems to regulate voltage and in energy storage systems. Additionally, parallel plate capacitors are utilized in various scientific instruments and equipment for precise measurements and experiments.

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