# Electron's energies and charge density in plate-capacitor

• late347
In summary, the electric potential energy of the electron while it begins its journey from the negative plate should be 0 because the potential is 0 at that location.
late347

## Homework Statement

a.)give voltage
b.)give the absolute value of charge density for plates

Presumably the absolute values of charge densities for plates is equal between the two of them.
An electron is accelerated in a homogenic electric field, inbetween two oppositely charged plates. Presumably the electron moves from one plate to the other plate.

The distance between plates = d = 0,01m

The starting velocity of the electron is assumed to be ##0= v_0##

The end velocity of the electron is ##1,9 * 10^7 m/s = v_1##

mass of electron ##m_e = 9,1*10^{-31} kg##

charge of electron ##q_e = -1,6*10^{-19} C##

## Homework Equations

electric potential energy ##E_p = q*V##, where V=potential q= charge
##E_k=0,5*mv^2##
potential V, distance d, strenght of field E ##V= E*d##
conservation of energy

## The Attempt at a Solution

[/B]
It could be assumed that minus plate = potential 0, I guess...
Well, in conventional gravitational potential energy and kinetic energy type of problems the good idea would be to try to use the conservation of energy principle.

from the electron's movement we know (I think?)
that:

E_{k0} =0 joules
E_{k1} = can be calculated with the given values

E_{p0}= this must be at the value of 0 joules if the formulas are true, because I suppose the electron does start from rest from the position of the negatively charged plate towards the positively charged plate. We can assume that at at the negative plate, the potential V=0

##E_{p0} = q_e * V##, where we can see that V=0, and we know that q_e = some negative charge
so the ##E_{p0}## = 0 if that formula is true... (because mathematically qV=0, when V or q are 0)

I calculated that the final kinetic energy ##E_{k1} = 1,64255* 10^{-16}## Joules, by using the kinetic energy formula E=0,5*mv^2

conservation of energy principle
##E_{k1} + E_{p1} =E_{k0} + E_{p0}##
##E_{k0}=0##
##E_{p0}=0##
<=>
##E_{k1} + E_{p1} =0##
<=>
##E_{k1} = -E_{p1}##

I'm little bit uncertain if this is above mentioned portion is true for the relationship between electron's kinetic energy and the electric potential energy of that same electron?My teacher talked about another formula which could be useful at this stage of the problem
##ΔE_{p}= q*ΔV= qU## where U = voltage

if that formula is true then it would seem that this follows:
##(E_{p1}-0)= q* (V_1-0)##
##(-1,64255*10^{-16} J = -1,6*10^{-19}C * V_1##

therefore essentially
voltage U = 1026,5937 VoltsI think the second part b.) was more confusing for me a little bit, especially the negative sign and positive sign issues.

Supposedly ΔV = U = voltage
then there should be a formula such as:
ΔV = -Ed
That formula describes change of potential in the same line as the electric field lines(?). E is the electric field strength and d is the distance moved.

if the voltage truly is about 1000Volts then:
1000V= -E*d
100000 V/m = -E
E= -100 000 V/m

(question: why is the electric field strength negative value at this point? now I am confused as heck)

With the Electric field strength one can use the formula to find out the charge density σ
E= σ/ε_{0}

=>

σ= - 8,85 *10^{-7} C/(m^2)

(question, again the negative sign is a little bit confusing at this stage
although admittedly one of the plates should have a negative value for the chage density and the other one should have the positive value)

so that essentially I think the idea was to calculate absolute value of charge density |σ|

But I think that those should be just about the correct answers.
For b.) the more accurate answer from my calculation without the rounding of variables inbetween calculation was something more like

##σ= -9,09 * 10^{-7} \frac{C}{m^2}##

late347 said:
I'm little bit uncertain if this is above mentioned portion is true for the relationship between electron's kinetic energy and the electric potential energy of that same electron?
What else?
late347 said:
voltage U = 1026,5937 Volts
Correct.
late347 said:
(question: why is the electric field strength negative value at this point? now I am confused as heck)
That just depends on the direction you look at (from minus to plus or from plus to minus). It doesn't matter.
late347 said:
although admittedly one of the plates should have a negative value for the chage density and the other one should have the positive value)
Right.
late347 said:
##σ= -9,09 * 10^{-7} \frac{C}{m^2}##
Right apart from the sign - the absolute value is positive.

late347
mfb said:
What else?Correct.That just depends on the direction you look at (from minus to plus or from plus to minus). It doesn't matter.
Right.
Right apart from the sign - the absolute value is positive.

I was initially confused a little bit about the idea that my teacher had.

E_kin1= -E_p1

But I managed to draw out that same conclusion from using the conservation of energy formula so it suddenly made sense again...

Was I correct also that the electric potential energy of the electron while it begins its journey from the negative plate, should be 0 because... the potential is 0 at that location ?

Well... at any rate you have to pick a zero point for the potential somewhere probably one of the charged plates (usually the negative one)

late347 said:
Was I correct also that the electric potential energy of the electron while it begins its journey from the negative plate, should be 0 because... the potential is 0 at that location ?
The potential energy is arbitrary, you can set it to whatever you want. Only potential differences between points have a physical meaning.

late347

## 1. What is a plate-capacitor?

A plate-capacitor is a device consisting of two conductive plates separated by a dielectric material that can store electrical energy by creating an electric field between the plates. It is commonly used in electronic circuits to store and release energy.

## 2. What is the relationship between electron's energy and charge density in a plate-capacitor?

The energy of an electron in a plate-capacitor is directly related to the charge density, which is the amount of charge per unit area on the plates. As the charge density increases, the energy of the electrons also increases.

## 3. How are the energies and charge density of electrons affected when the distance between the plates is changed?

When the distance between the plates of a plate-capacitor is increased, the charge density decreases and the energy of the electrons decreases as well. This is because the electric field between the plates becomes weaker as the distance increases.

## 4. What happens to the energies and charge density of electrons when a voltage is applied to a plate-capacitor?

When a voltage is applied to a plate-capacitor, the charge density and energy of the electrons increase. This is because the voltage creates a stronger electric field between the plates, causing more electrons to be attracted to one plate and repelled from the other.

## 5. Can the energy and charge density of electrons in a plate-capacitor be manipulated?

Yes, the energy and charge density of electrons in a plate-capacitor can be manipulated by changing the distance between the plates, the voltage applied, or the type of dielectric material used. These factors all affect the electric field and therefore, the energy and charge density of the electrons in the capacitor.

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