Point charge between two parallel plates (capacitor)

[Rod Alexei]

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.

The Attempt at a Solution

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

(c) W= /_\K
I do not know what to do next.
(d) I also do not know where to begin.

 

[cnh1995]

What is the initial distance between the charge and the positive plate? Isn't that mentioned?

 

[Rod Alexei]

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.

 

[cnh1995]

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?

 

[Rod Alexei]

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.

 

[cnh1995]

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?

 

[Rod Alexei]

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 ?

 

[cnh1995]

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?

 

[Rod Alexei]

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

 

[gneill]

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 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?

 

[Rod Alexei]

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 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

 

[cnh1995]

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?

 

[Rod Alexei]

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

 

[gneill]

I might suggest that given the requested items, an energy approach might be expedient... You have a constant acceleration (just like gravity).

 

[Rod Alexei]

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

 

[cnh1995]

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.

 

[Rod Alexei]

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?

 

[gneill]

It often helps to re-draw the problem so that it reminds you of a familiar scenario. A simple rotation of the image gives:

 

[cnh1995]

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.

 

[Rod Alexei]

So I just have to add .3 m to the value that I got for the max height correct? Thank you very much

 

[cnh1995]

So I just have to add .3 m to the value that I got for the max height correct? Thank you very much

Yes.

 

[Rod Alexei]

For the final question, the particle has the largest speed right before it comes into contact with the positively charged plate?

 

[cnh1995]

For the final question, the particle has the largest speed right before it comes into contact with the positively charged plate?

Yes.

 

[Rod Alexei]

Here is my answer to number 3:
sin^2 45 deg/(2x0.9m/s^2) +0.3 = 0.58 m

 

[cnh1995]

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.

 

[Rod Alexei]

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?

 

[cnh1995]

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.