Point charge between two parallel plates (capacitor)

In summary, a fixed parallel plate capacitor with infinite plate length and a capacitance of 5.0 F is charged with 3 C and -3C on its plates. A point charge with a mass of 0.80 kg and an initial speed of 1.0 m/s is projected at an initial distance of 0.3 m from the positive plate. The potential difference between the plates is 0.6 V. The nearest distance between the point charge and the negatively charged plate during motion cannot be determined without knowing the initial distance between the charge and the positive plate. The largest speed of the point charge during the motion can be calculated using SUVAT equations, but the necessary variables are not provided. An energy approach could
  • #1
Rod Alexei
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1. Homework Statement
A fixed parallel plate capacitor is charged. A point charge is starting to move with an initial speed 1.0 m/s at shown initial position. The capacitance is 5.0 F. The mass of the point charge is 0.80 kg. Assume that the plate length is infinite. No gravity. The initial distance between the charge and the positive plate is 0.3 m. The charges of the two plates are 3 C and -3C.

Homework Equations


(a) What is the potential difference between the two plates?
(b) Draw the trajectory of the point charge on the figure given below.
(c) Find the nearest distance between the point charge and the negatively charged plate during motion.
(d) Find the largest speed of the point charge during the motion.
p6.png


The Attempt at a Solution


(a) /_\ (delta) v = Q/C= 3/5 = 0.6 V
(b)
pg answr b.png

(c) W= /_\K
I do not know what to do next.
(d) I also do not know where to begin.
 
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  • #2
What is the initial distance between the charge and the positive plate? Isn't that mentioned?
 
  • #3
cnh1995 said:
What is the initial distance between the charge and the positive plate? Isn't that mentioned?
It wasn't mentioned. This is a problem in our midterm that I failed to answer.
 
  • #4
Rod Alexei said:
It wasn't mentioned. This is a problem in our midterm that I failed to answer.
Well, I believe we can't answer part c) without that.
Ok. What do you know about the force acting on the charge? What is the origin of that force?
 
  • #5
cnh1995 said:
Well, I believe we can't answer part c) without that.
Ok. What do you know about the force acting on the charge? What is the origin of that force?
I'm sorry I just remembered it was 0.3 meters. My bad. The force on the charge would be constant I guess anywhere since the electric fields between the plates are constant.
 
  • #6
Rod Alexei said:
The force on the charge would be constant I guess anywhere since the electric fields between the plates are constant.
Yes. What is the magnitude of the electric field? You have voltage between the platesand its physical dimensions. What is the force on a charge placed in an electric field? What will be its direction in this case?
 
  • #7
cnh1995 said:
Yes. What is the magnitude of the electric field? You have voltage between the platesand its physical dimensions. What is the force on a charge placed in an electric field? What will be its direction in this case?
The magnitude of F on point charge = Eq. The direction would be opposite the electric field since charge is negative ?
 
  • #8
Rod Alexei said:
The magnitude of F on point charge = Eq. The direction would be opposite the electric field since charge is negative ?
Right. What is the magnitude of the electric field in this case?
 
  • #9
cnh1995 said:
Right. What is the magnitude of the electric field in this case
I can't use coulomb's law right since F is constant? SO to get magnitude of E I have to use a different formula:
E=/_\ (delta) V/d where d is distance between the two parallel plates.
E=0.6 V (from no.1) / 1.0 m = 0.6 N/C
 

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  • #10
Rod Alexei said:
A fixed parallel plate capacitor is charged. A point charge is starting to move with an initial speed 1.0 m/s at shown initial position. The capacitance is 5.0 F. The mass of the point charge is 0.80 kg. Assume that the plate length is infinite. No gravity. The initial distance between the charge and the positive plate is 0.3 m.

For clarification, how is the charge on the capacitor specified? I see 3.0 and -3.0 indicated next to the plates, but no units are given. For all I know they might be nano coulombs or potentials in kV or charge densities.

A capacitor with infinite plate length and a finite capacitance is quite an achievement o_O I presume you mean to say that the plates are large enough that edge effects can be ignored so that the electric field between them is uniform?
 
  • #11
gneill said:
For clarification, how is the charge on the capacitor specified? I see 3.0 and -3.0 indicated next to the plates, but no units are given. For all I know they might be nano coulombs or potentials in kV or charge densities.

A capacitor with infinite plate length and a finite capacitance is quite an achievement o_O I presume you mean to say that the plates are large enough that edge effects can be ignored so that the electric field between them is uniform?
Yes. They are in COulombs
 
  • #12
Your electric field calculation is correct. Now, the particle is projected with some angle with respect to the plates. So it will have two components of velocity: 1) parallel to the plates and 2) perpendicular to the plates.
Which component would be affected by the force? How would you apply SUVAT equations (kinematic equations) to these components?
 
  • #13
cnh1995 said:
Your electric field calculation is correct. Now, the particle is projected with some angle with respect to the plates. So it will have two components of velocity: 1) parallel to the plates and 2) perpendicular to the plates.
Which component would be affected by the force? How would you apply SUVAT equations (kinematic equations) to these components?
The force will only affect the y component of the velocity. I believe if I change the way I view the plates (horizontally) and then draw the electric field lines I can think of it as gravity.
so F = 0.6 N/C (-1.2 C)
F = -0.72 N
so a = -0.72/.80 = -0.9 m/s
I think I should use the
vf^2=vi^2 + 2a (xf-xi)
But I don't know two variables, vf and xf (What I want I guess)
So getting the components of vi (initial velocity)
I got vxi = cos 315 deg from positive x-axis (1) m/s
vyi=sin 315 deg from positive x - axis (1) m/s
With the x-axis parallel to the original figure (vertical orientation) and y is perpendicular
 
  • #14
I might suggest that given the requested items, an energy approach might be expedient... You have a constant acceleration (just like gravity).
 
  • #15
gneill said:
I might suggest that given the requested items, an energy approach might be expedient... You have a constant acceleration (just like gravity).
Yes but I am quite uncomfortable with energy since we have not thoroughly discussed it. (we skipped most of the chapters).
I'd like to know how to start
 
  • #16
Rod Alexei said:
The force will only affect the y component of the velocity. I believe if I change the way I view the plates (horizontally) and then draw the electric field lines I can think of it as gravity.
so F = 0.6 N/C (-1.2 C)
F = -0.72 N
so a = -0.72/.80 = -0.9 m/s
I think I should use the
vf^2=vi^2 + 2a (xf-xi)
But I don't know two variables, vf and xf (What I want I guess)
So getting the components of vi (initial velocity)
I got vxi = cos 315 deg from positive x-axis (1) m/s
vyi=sin 315 deg from positive x - axis (1) m/s
With the x-axis parallel to the original figure (vertical orientation) and y is perpendicular
Ok.
Have you studied projectile motion? This is an example of projectile motion and you can use all its formulae by replacing gravitational acceleration 'g' with the acceleration you just calculated.
 
  • #17
cnh1995 said:
Ok.
Have you studied projectile motion? This is an example of projectile motion and you can use all its formulae by replacing gravitational acceleration 'g' with the acceleration you just calculated.
Yes so can I use the formula that doesn't involve time?
SO is the nearest distance (y direction) at the vertex of the parabola?
So I can just plug into the formula for the maximum height?
 
  • #18
It often helps to re-draw the problem so that it reminds you of a familiar scenario. A simple rotation of the image gives:

upload_2016-11-2_12-39-4.png
 
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  • #19
Rod Alexei said:
Yes so can I use the formula that doesn't involve time?
SO is the nearest distance (y direction) at the vertex of the parabola?
So I can just plug into the formula for the maximum height?
Yes. But keep in mind that the particle started at 0.3m from the +ve plate. Don't forget to take it into account while calculating maximum height. Refer gneill's modified diagram.
 
  • #20
So I just have to add .3 m to the value that I got for the max height correct? Thank you very much
 
  • #21
Rod Alexei said:
So I just have to add .3 m to the value that I got for the max height correct? Thank you very much
Yes.
 
  • #22
For the final question, the particle has the largest speed right before it comes into contact with the positively charged plate?
 
  • #23
Rod Alexei said:
For the final question, the particle has the largest speed right before it comes into contact with the positively charged plate?
Yes.
 
  • #24
Here is my answer to number 3:
sin^2 45 deg/(2x0.9m/s^2) +0.3 = 0.58 m
 
  • #25
Rod Alexei said:
Here is my answer to number 3:
sin^2 45 deg/(2x0.9m/s^2) +0.3 = 0.58 m
You are asked to find the nearest distance between the charge and the negative plate.
 
  • #26
cnh1995 said:
You are asked to find the nearest distance between the charge and the negative plate.
Yes thank you very much I found the answer now. I got 0.42 m. Thank! For number four can I use a kinematics formula as well?
 
  • #27
Rod Alexei said:
Yes thank you very much I found the answer now. I got 0.42 m. Thank! For number four can I use a kinematics formula as well?
Yes.
 

1. What is a point charge between two parallel plates?

A point charge between two parallel plates refers to a scenario in electrostatics where a single charged particle is placed between two conducting plates that are parallel to each other. This setup is commonly known as a parallel plate capacitor.

2. What is the purpose of a point charge between two parallel plates?

The purpose of a point charge between two parallel plates is to create an electric field between the plates. This electric field can be used to store electrical energy and is the basis for many electronic devices such as capacitors and batteries.

3. How is the electric field strength affected by the distance between the plates?

The electric field strength between the plates is directly proportional to the distance between the plates. As the distance between the plates increases, the electric field strength decreases.

4. What is the formula for calculating the electric field between two parallel plates?

The formula for calculating the electric field between two parallel plates is E = Q/εA, where E is the electric field strength, Q is the charge on the plates, ε is the permittivity of the material between the plates, and A is the area of the plates.

5. How does the charge on the plates affect the electric field between them?

The charge on the plates has a direct impact on the electric field between them. As the charge on the plates increases, the electric field strength also increases. Conversely, decreasing the charge on the plates will result in a decrease in the electric field strength.

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