Determining the Charge on Parallel Plates

[Brandone]

Homework Statement

An electron starts from one plate of a charged closely spaced (vertical) parallel plate arrangement with a velocity of 1.63x10⁴ m/s to the right. Its speed on reaching the other plate, 2.10 cm away, is 4.15x10⁴ m/s.
...
If the plates are square with an edge length of 25.4 cm, determine the charge on each.

Given and Known:
v_i = 1.63x10⁴
v_f = 4.15x10⁴
d = 2.10 cm
A = (25.4 cm)² = (0.0645 m²)
m_e = 9.109x10^-31 kg
q_e = 1.602x10^-19 C
k = 9.00 x 10⁹ Nm²/C²

Homework Equations

a = (v_f-v_i) / (t_f-t_i)

F = ma

E = F_q / q

Q = EA / 4[pi]k

The Attempt at a Solution

First, I used:

F = ma, and
E = F_q / q

And got:

E = ma / q
E = (2.52x10⁴-N/C) (9.109x10^-31 kg) / (1.602x10^-19 C)
E = 6.823 x 10^-6 N/C

Then,

Q = EA/4[Pi]k
Q = (6.823 x 10^-6 N/C) (0.0645 m²) / 4[Pi](9.00x10⁹ Nm²/C²
Q = 3.89 x 10⁴ C

So, this does not agree with the answer key's 1.13x10^-13 C, but it seems like I'm making sensible steps toward the answer.

Guidance, please?

 

[rl.bhat]

Using the kinematic equation vf^2 - vi^2 = 2*a*s, find a.
F = ma = E*q

 

[vorcil]

Homework Statement

An electron starts from one plate of a charged closely spaced (vertical) parallel plate arrangement with a velocity of 1.63x10⁴ m/s to the right. Its speed on reaching the other plate, 2.10 cm away, is 4.15x10⁴ m/s.
...
If the plates are square with an edge length of 25.4 cm, determine the charge on each.

Given and Known:
v_i = 1.63x10⁴
v_f = 4.15x10⁴
d = 2.10 cm
A = (25.4 cm)² = (0.0645 m²)
m_e = 9.109x10^-31 kg
q_e = 1.602x10^-19 C
k = 9.00 x 10⁹ Nm²/C²

Homework Equations

a = (v_f-v_i) / (t_f-t_i)

F = ma

E = F_q / q

Q = EA / 4[pi]k

The Attempt at a Solution

First, I used:

F = ma, and
E = F_q / q

And got:

E = ma / q
E = (2.52x10⁴-N/C) (9.109x10^-31 kg) / (1.602x10^-19 C)
E = 6.823 x 10^-6 N/C

Then,

Q = EA/4[Pi]k
Q = (6.823 x 10^-6 N/C) (0.0645 m²) / 4[Pi](9.00x10⁹ Nm²/C²
Q = 3.89 x 10⁴ C

So, this does not agree with the answer key's 1.13x10^-13 C, but it seems like I'm making sensible steps toward the answer.

Guidance, please?

Use the kinematic equations $$V_f = V_i + a*t$$
and $$d = \frac{ V_f + V_i}{2}*t$$
to solve for the acceleration the electron has between these two plates

use F = ma to solve for the force the electron experiences,

then part of Coulomb's law relating to the super positioning of electric fields states that
F = QE, where F is the force the particle experiences, Q is the charge of the particle and E is the electric field strength,

using that you should be able to solve a number for the electric field,

however the next bit may be a little tricky, you will need to derive an expression for the electric field between two parallel plates,

to do this, you will need to apply Gauss's law,

$$\oint_S \mathbf{E}.d\mathbf{a} = \frac{1}{\epsilon_0}Q_{enc}$$

I'm not sure if I'm allowed to show you how to derive the electric field between two parallel plates (forum rules) but I can tell you that you'll need to draw a Gaussian pillbow, that extends above and bellow the plate,

in the equation above, your enclosed charge $$Q_{enc} = \sigma A$$ where sigma is the surface charge density of the plate, and A is the area of the lid of the pill bow,

 

[vorcil]

I'm not sure If I am allowed to show Brandone how to get an expression for the electric field between two plates,

It's not an easy thing to derive for an introductory physics forum

so I'm going to show him how I did it, if this is wrong, or against the forum rules(within reason) the mods can delete this post.

------

Gauss's law:

$$\oint \mathbf{E} . d\mathbf{a} = \frac{1}{\epsilon_0}Q$$

assuming the electric field is constant, (i.e plates stay at their charge)
also this is the electric field from a single plate

$$E \int d\mathbf{a}$$

= $$2A|\mathbf{E}|$$

- going back to the original equation

$$2AE = \frac{1}{\epsilon_0}Qenc$$

Qenc is the total charge on the plate, = $$\sigma A$$
where A is the Area of the plate, and Sigma is the surface charge density of the plate

giving me the equation

$$2AE = \frac{1}{\epsilon_0}\sigma A$$
the As cancel out, and I can re arrange the equation as,

$$2E\epsilon_0 = \sigma$$

the surface charge density is the charge on the plate/area $$\sigma = \frac{Q}{A}$$

substituting

$$2E\epsilon_0 = \frac{Q}{A}$$

$$Q = 2E \epsilon_0 A$$ taking out the 2 by taking into account this is the field for one plate, note two plates I get

$$Q = E \epsilon_0 A$$

This is the equation for the electric field between two plates with a charge difference of Q (I think)

 

[vorcil]

http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter23/Chapter23.html

23.11 shows how the field is derived in a similar way, (easier to understand way if you ask me, compared to my Glaw method)

having the electric field of one sheet to be $$E = \frac{\sigma}{2\epsilon_0}$$
and between two sheets of opposite charge, you get E = Etop+Ebottom so the two cancels out
giving the same result I had.

 

[Brandone]

I don't know from what dusty corner of my brain I pulled that equation for acceleration. Ugh... silly kinematics.

Vorcil, the issue with that, is that I'm to show all work, and there is no calculus in this classical course. If I work out the derivative on the homework, my professor would be rather curious, I'm afraid.

I had thought Q=EA/4[pi]k is the charge on a parallel plate configuration. Is that not it? I'm looking for total charge--not charge density.

 

[rl.bhat]

I had thought Q=EA/4[pi]k

Your formula is correct.

Check your acceleration and electric field.