- #1

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## Homework Statement

Tasks to do:

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...

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}##