# Electric field between two parallel plates

• meher4real
C : CapacitanceV : Voltageq : electric chargee0 : permitivity of vacuumA : plate aread : distance between platesIn summary,An electron beam with velocity vector v = (0; 0.6x10^8 ;0) m.s enters between two oppositely charged plates parallel to the xz plane.The areal charge density on the plates is 3.6x10^19 C.The voltage for the plate spacing of 25mm is 9.5 V.

#### meher4real

Homework Statement
Can't find the solution for this problem.
Relevant Equations
Wish i do know the equation.
[New poster has been reminded to show their best efforts to work the schoolwork problem when starting a homework thread]

My question is : An electron beam with velocity vector v = (0; 0.6x10^8 ;0) m.s enters between two oppositely charged plates parallel to the xz plane.
- How large is the areal charge density on the plates when the electric field between the plates stop the beam at a distance of 10mm ?
- Determine the voltage for the plate spacing of 25mm ?

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Please tell us what level course this is for and indicate your thoughts about how to solve this.

hutchphd said:
Please tell us what level course this is for and indicate your thoughts about how to solve this.
I study Mining of Mineral Resources (Geology-Master Degree)
- I think that the question is tricky because the majority of electric field type of questions i found on the internet mention the q charge and they're different from this problem.

Is this an assigned problem for a particular class? What tools can you bring? For instance what does the electric field look like between two charged parallel conducting plates?

I've tried with the equation a=q/EM => a=qU/md
But none of U or q is available.

meher4real said:
I study Mining of Mineral Resources (Geology-Master Degree)
- I think that the question is tricky because the majority of electric field type of questions i found on the internet mention the q charge and they're different from this problem.
What is the q charge for an electron? I am sure you can find that on the internet.

Maybe what you found on the internet is different, but the question is not tricky. Read the problem carefully. The electric field stops the beam. Make a drawing showing the electric field lines and the velocity of a single moving electron in the beam. Whatever one electron does, all the electrons in the beam do.

The real trick is in asking the right questions that will lead you to the answer. Here are two to get you started.
What can you say looking at this picture? What must be true if an electron, moving at 0.6×108, stops over a distance of 10 mm? I will stop here and leave asking the remaining questions to you. There is no magic formula that will give you the answer. You have to put it together (with help).

• rudransh verma and hutchphd
What is the direction of the field.?

Thank you guys for your interest but you have to know that I'm not a physicist, I'm a geologist.
Before i come here i did some research to find a similar problems, all of them them have the electron leave the parallel plates.
The question is not attached with a picture to know how it works. The problem is exactly what i mentioned.

Geologists are allowed to learn physics!
Who is posing this question? It is rather specific, and requires knowledge of the physics.

hutchphd said:
Geologists are allowed to learn physics!
Who is posing this question? It is rather specific, and requires knowledge of the physics.
I meant that you don't have to learn everything.
It's a homework, I've finished three problems today (2 about waves and 1 optic) but this one is difficult for"me" (geologist).

Maybe I can remind you the formula $$E=\frac{\sigma}{\epsilon_0}$$ where E is the electric field between the plates and ##\sigma## the surface (or areal) charge density.

Also do you remember the work-energy theorem?

• meher4real
meher4real said:
Thank you guys for your interest but you have to know that I'm not a physicist, I'm a geologist.
Before i come here i did some research to find a similar problems, all of them them have the electron leave the parallel plates.
The question is not attached with a picture to know how it works. The problem is exactly what i mentioned.
meher4real said:
I meant that you don't have to learn everything.
It's a homework, I've finished three problems today (2 about waves and 1 optic) but this one is difficult for"me" (geologist).
Picking random formulas and solutions off of the Internet is not a good strategy for learning.

The electron enters a region with a uniform electric field, and is subject to a constant retarding force. What is the equation for that force given the electric field in that region?

And using the mass of the electron, what is the deceleration of the electron in that region? How long does it take for it to stop?

• meher4real
What do you think about using Capacitance formula ?
C = q/V = e0x(A/d)
C : Capacitance
V : Voltage
q : electric charge
e0 : permitivity of vacuum
A : plate area
d : distance between plates

meher4real said:
What do you think about using Capacitance formula ?
C = q/V = e0x(A/d)
Another tip would be for you to check out the LaTeX tutorial link in the lower left of the Edit window. It helps you to post math equations in online forums like the PF. $$C = \frac{q}{V} = e_0\frac{A}{d}$$

meher4real said:
What do you think about using Capacitance formula ?
C = q/V = e0x(A/d)
I don't think that would be helpful. Let's simplify the problem, you have an electron with initial velocity ##v_0=0.6\times 10^8m/s## that enters an electric field and stops after traveling distance ##d=10mm##. Can you determine the strength of the electric field E (i suppose the mass and charge of the electron is given too).

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Apply Lorentz's Law ?
F = qv x B

We don't have a magnetic field B present in this problem. What is the formula for the force that involves the electric field E?

• berkeman
There are two formulas :
E=F/q=kq/r^2 => F=Eq ?
V=U/q=W/q=Ed=kq/r ?

Well ok, let's take F=Eq. So that is the force from the electric field on the electron. What is the work of this force over a distance of d=10mm?

It might be the case that you haven't been taught the concept of the work of the force, in which case we ll have to do this via deceleration and kinematics equations as berkeman suggests at post #12.

E=14.4x10^-6 right ?

Er, i don't know, all i was planning to do was to guide you to find the right equation and then you could plug in the numbers to find the exact value.
Can you tell us how did you arrive at that number? All the intermediate steps?

Understood.
k = Coulomb's constant = 9x10^9
q = electron charge = 1.60x10^-19
E = F/q = kq/d^2 = 14.4x10^-6

The formula E=kq/d^2 is not correct for this problem. That is the electric field from a particle having charge q at distance d from the particle. In this problem the electric field is due to the surface charge density of the plates, not due to the electron.

I asked you a question about the work of the force F=Eq (E is our unknown here which we will try to find from the other data given and q is the charge of electron). Can you answer that question?

• meher4real
You have a constant force F=Eq that acts over a distance of d(=10mm). What is the work of this force. The next step will be to equate this work with the change in kinetic energy of the electron over the distance of d=10mm. And from that equation you ll find E.

Have you been taught the concept of work and the work-energy theorem? Please at least answer this question...

Delta2 said:
Have you been taught the concept of work and the work-energy theorem? Please at least answer this question...
I have no clue about Electromagnetism.

F = kqq/r^2

Work and Work-Energy theorem is Classical Mechanics.
No Coulomb's law won't help you here because as I said before the electric field is not due to point particles but due to the surface charge density of the parallel plates.

Let me explain to you my situation here.
I used to study physics in french (my last course was four years ago), this is my first time doing it in English, plus i used to work for three years on marketing, this year I've decided to complete my education and quit my job so i can get a master degree in Geology as a foreign student.
It's a challenging stage for me and I'm motivated to pass it.
(Hope you understand why I'm struggling here)

Ok I see now, in French and 4 years ago, that explains it :D.

Unfortunately I don't know French (except a few words hehe) so I don't know how Work and Work-Energy theorem is in French.

We ll try to do this with the kinematic approach. Do you recall the equations ##d=v_0t-\frac{1}{2}at^2##, ##v=v_0-at##?

• • Steve4Physics and meher4real
Thank you for understanding Of course, but how it can work with Electromagnetism ?

For this problem we don't use a lot from Classical Electromagnetism, just that ##F=Eq## and that ##E=\frac{\sigma}{\epsilon_0}##, the rest is classical mechanics stuff.
Also for the second part you have to use the formula ##V=E\cdot l## where l is the spacing of the two plates.

If the force is ##F=Eq## what is the acceleration ##a## from the electric field on the electron?