# Homework Help: You all are such a blessing I need help.

1. Aug 26, 2006

### duki

Hey everyone,

I'm 18 and a freshman taking Physics II (electr.). I am in desperate need of help Without exagerating, I have been working on 3 problems for the past hour. If someone could please help me I would appreciate it so much...

Here's one of my problems:

Book: Cutnell & Johnson Physics 7th Edition
pp.569 #38

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is o=1.8 x 10^-7 C/m^2, and the plates are separated by a distance of 1.5 x 10^-2m. How fast is the electron moving just before it reaches the positive plate?

My Answer: 1.43 x 10^26 m/s

I don't think my answer is accurate. If there was a way to find the initial velocity without first having the final velocity, I think I could figure out one of the others I have.

Anyways, Thanks for any help guys!

Last edited: Aug 26, 2006
2. Aug 26, 2006

### Staff: Mentor

Don't just give your answer, show how you got your answer.

As far as how to "find the initial velocity", I have no idea what you mean. The initial velocity is given as zero (it was "released from rest").

3. Aug 26, 2006

### duki

Well here's what I did...

F = (8.99 x 10^9)(1.8 x 10^-7)(1.8 x 10^-7) / (1.5 x 10^-2)^2 = 1.3N

Then...

Ax = 1.3N / 9.11 x 10^-27 = 1.43 x 10^26 m/s

And the velocity is for a different question. I'll post it in a different thread. I'm lost on this stuff, I appologize.

4. Aug 26, 2006

### leright

ok, do you know how to find the electric field in between the plates of a parallel plate capacitor? You can determine the formula by finding the e-field due to a infinite sheet of charge using gauss's law, and then the e-field in between the plates of the capacitor is obviously 2 times the e-field due to an infinte sheet. The key is that the electric field due to an infinite sheet is a constant, and therefore the e-field in between the plates of a parallel plate cap is also a constant.

Now, when you have the e-field in between the plates of the capacitor you can find the force on the electron, by definition of the electric field. F = qE.

If you have the force on the electron you can find the acceleration using Newton's second law, right? You can look the mass of an electron up in your book, probably.

Now, you know the acceleration of the electron. Now you need to know the time it takes the electron to travel from the negative plate to the positive plate. You can do this quite easily using the kinematic equations. Do you know these, or do you know how to derive them?

acceleration = A
velocity = At + (initial velocity)
displacement = 0.5At^2 + (initial velocity)*t + (initial displacement)

You know the displacement and the acceleration, and you know the initial velocity and initial displacement are both zero. From the third kinematic equation you can determine the time it takes for the electron to get from the neg plate to the pos plate. Just multiply the acceleration by this time to get the velocity when the electron reaches the positive plate.

Does this help?

And yes, I would agree that your answer is not accurate. :p The speed of light is ~300,000,000. Your speed is much higher than this, and nothing can exceed the speed of light.

Last edited: Aug 26, 2006
5. Aug 27, 2006

### Staff: Mentor

Unfortunately PF crashed just as I was answering your post yesterday. But it looks like leright has given you sound advice. In anycase, my comments:
Looks to me like you are applying Coulomb's law here, which is used to find the force between two charges. It's not relevant here. (You need to find the field between the plates, then the force experienced by the electron due to that field.)
Here it looks like you applied Newton's 2nd law to find the acceleration, then just called it velocity.

Follow leright's excellent advice to solve this properly.

6. Aug 27, 2006

### duki

In the problem it has 1.8 x 10^-7 C/m^2 ... What is C/m^2? Coulombs per square meter? How do I input that into Gauss's Law? The answer I'm getting for my e-field is the following:

E = 1.8 x 10^-7 / 1.5 x 10^-2 = 1.2 x 10^-05 (if multiplied by two = 2.4 x 10^-05)

After that, what is q in F = qE? 1.6 x 10^-19?

I appologize if I'm asking dumb questions...

Last edited: Aug 27, 2006
7. Aug 27, 2006

### leright

C/m^2 is the surface charge density of the capacitor plates. Yes, it is coulombs per square meter. Do you know what gauss's law states? I will make it easy for you and I will tell you that the e-field due to an infinte sheet of charge is (surface charge density)/(2*(permittivity of free space)). The e-field due to one plate would then be (1.8x10^-7)/(2*(8.854x10^-12)) = 10,164.90 N/C. The total e-field in between the plates would be 2*(10,164.90) = 20,329.80 N/C. Clear as mud?

In F = qE, q is the charge placed in the electric field. In this case, it is the charge of one electron, which is -1.6x10^-19 C. So the force is (-1.6x10^-19)*(20,329.80) = -3.25x10^-15 N. The negative sign means that the force is in the opposite direction of the e-field, and the e-field points from positve plate to negative plate. Therefore, the force is from negative plate to positive plate. This is intuitively pleasing, correct?

Now, we can find the acceleration of the electron in this e-field as it travels from negative plate to positive plate using Newton's second law, f=ma. a = (3.25x10^-15)/(9.1x10^-31) = 3.57x10^15 m/s^2

Now we need to find the time it takes for the electron to travel from negative plate to positive plate using the following kinematic equation.

displacement = 0.5At^2 + (initial velocity)*t + (initial displacement).

We know that the initial velocity and initial displacement are both zero. Do you know why this is? Therefore, displacement = 0.5At^2. We know the displacement is the distance between the capacitor plates and we just calculated the acceleration, so the time is simply t = sqrt((2*(1.5x10^-2))/(3.57x10^15)) = 2.9x10^-9 s.

Finally, the velocity of the electron by the time it reaches the positive plate is simply V = (time it takes for electron to reach the positive plate)*(acceleration of electron) = (2.9x10^-9)*(3.57x10^15) = 10,353,000 m/s

Clear as mud?

As a side note, are you aware how to apply Gauss's law to determine the e-field of a sheet of charge? I recommend you try this yourself since it is an important skill to have and I am sure you will be tested on it. Just state in your mind what Gauss's law says and then see if you can solve it.

8. Aug 27, 2006

### duki

Wow, thanks. It's making a little more sence, but I'm going to look up this stuff in my book and write it down in my notes. I really appreciate your help. Just a few questions if you don't mind...

To find the electric field you always have to multiply the permittivity of free space by 2?

in the displacement, you always divide acceleration and time^2 by 2? (.5At^2)?

No, I do not know why the iv and id are zero. I understand that it is because the electron is released from rest, but could you clarify?

Thanks a bunch!! I'll get to writing notes and check back on my questions.

9. Aug 27, 2006

### leright

The e-field equation I gave you is only for infinite sheets of charge! It is not the same equation for every charge distribution.

That is the expression for the displacement in cases where the acceleration is constant and initial velocity and initial displacement are both zero. The velocity as a function of time is found by taking the integral of acceleration wrt time and the displacement as a function of time is found by taking the integral of velocity wrt time. The expression I gave for displacement is what you obtain when you do this. Do you understand why velocity is the integral of acceleration and the displacement is the integral of velocity?

The initial velocity is zero because the electron starts at the negative plate AT REST. The initial displacement can be made zero if you define the negative plate as the point where displacement = 0.

10. Aug 27, 2006

### duki

No.
What is wrt?

11. Aug 27, 2006

### leright

with respect to.

and I just explained why Vi and Di are zero.

12. Aug 28, 2006

### duki

Oh, ok. Thanks.

btw. You helped me so much. Apparently I wasn't the only one having a hard time in class. I had atleast 4 people asking me about how I did my work. Thank you so much. :rofl:

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