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Electron's energies and charge density in plate-capacitor

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data

    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##

    2. Relevant 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
    3. The attempt at a solution

    It could be assumed that minus plate = potential 0, I guess...
    electron's kinetic and potential electric energies.jpg


    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 Volts


    I 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 im 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}##
     
  2. jcsd
  3. Dec 9, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
  4. Dec 9, 2016 #3
    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)
     
  5. Dec 9, 2016 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The potential energy is arbitrary, you can set it to whatever you want. Only potential differences between points have a physical meaning.
     
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