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When does work=-electric potential and when does it = +

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  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    I have been trying to grasp the concept of work and electric potential energy inbetween two parallel plates with a uniform electric field and it is quite confusing. can someone explain when is work done = negative electric potential and when is it positive electric potential. i have been trying to figure this out for 3 days now please tell me straight forward.

    2. Relevant equations

    work.png work.png Think like the only thing given is change in voltage and the charge.
    3. The attempt at a solution
    I have looked throught textbook, online, this forums, and asked my professor. and no one has really helped me understand this concept. i know what potential energy ke and work is. i understand gravitational pe but not electrical. please someone explain.
     
  2. jcsd
  3. Dec 10, 2014 #2

    ehild

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  4. Dec 10, 2014 #3
    your answer makes no sense. your your change in voltage is vinitial-vfinal which is wrong your diagram confuses me u say it goes from A to B so i dont know about voltage change. i dont know if its a negative charge or positive either. when i try to use conservation of energy KEo+PEo=KEf+PEf it become -ΔKE=ΔPE ,W=ΔKE but when you plug in a proton for the charge u cannot get final velocity since u cannot sqrt a negative answer which leads me to think this is wrong and for this example W cannot equal negative change in potential energy. but it would be perfectly fine if we use a negative charge? someone please help me. would be nice if someone could just give me a right solid answer, i have prolly read my textbook over atleast 5+ times its sht. my notes say the opposite of what my txtbook says, my professor is confusing, and the answers i get are so vague sorry if i am a ***hole but im fed up spent so many days trying to figure this out
     
  5. Dec 10, 2014 #4

    ehild

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    Well, I write some more nonsense for you. I presume you can read ??? As you misread what I wrote in my post.
    That is what I have written in the other thread
    workpe-jpg.76387.jpg

    Your figures:

    sanhuy.JPG


    If you prefer using conservation of energy : The positive charge travels from A to B if UA>UB.
    KE(A)+PE(A)=KE(B)+PE(B). PE(A)=qUA, PE(B)=qUB. -ΔKE=ΔPE=q(UB-UA). As UB<UA, UB-UA is negative. What is the sign of ΔKE?
    Work-Energy theorem states that the change of KE equals to the work done. So is the work positive or negative?
     
  6. Dec 10, 2014 #5
    thing is if we are only given Δv = +2500v how do you know if Ua > Ub or vice versa. I understand in potential energy of gravity you can set the initial height to zero can you set the initial voltage to 0 or final to zero. if so which one if change in voltage is +2500v? sorry for being an ***hole like i said in my first post, but if i cant understand something and you can its probably my fault for being so dumb.
     
  7. Dec 10, 2014 #6

    ehild

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    If the particle goes from A to B the potential difference is UB-UA. You can set UA=0, but in case of positive Δv B is positive with respect to A. If A is positive with respect to B, the change of potential is negative.
    Some people use"voltage" for negative potential difference. Look at the problem wording carefully, or ask if voltage is mentioned. How the original problem was worded?
     
    Last edited: Dec 10, 2014
  8. Dec 10, 2014 #7
    so if the ΔV= +2500v like in a problem i am doing
    I set point A to 0 and point B to +2500 since it is going from A to B it is single charged proton and i have to find final velocity at point B
    so if i use
    KEo+PEo = KEf+PEf
    ½×M×Vo2 + q(Ua) = ½×M×Vf2 +q(Ub)
    0 + 0 = ½MVf2 + (1.6x10-2)(2500V)
    -½MVf2 = (1.6x10-2[/SUPc])(2500V)
    than Vf = √-2((1.6x10-2c)(2500V))

    which is impossible.

    my question tells use ΔV=2500v and "Motion of the positively-charged isotopes toward the right was initiated by a potential difference of 2500V on the two plates shown"
     
  9. Dec 10, 2014 #8
  10. Dec 10, 2014 #9

    ehild

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    Potential difference means only the magnitude here. The clue is "Motion of the positively-charged isotopes toward the right was initiated by a potential difference of 2500V"The positive charge is repelled by the positive plate so which plate is positive with respect to the other if the positive particle is driven to the right?

    Note the difference between potential difference and change of the potential. . Change of potential between an initial and final position is Ufinal-Uinitial. Is the final potential lower or higher than the initial one?

    You know the water flows in the direction of lower potential. The same with positive charges. "By themselves" they move towards the lower potential.
     
  11. Dec 10, 2014 #10
    so since the initial potential is high and protons flow to lower potential would the initial voltage be 2500v (Ua) or is that assumption invalid?
     
  12. Dec 10, 2014 #11

    ehild

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    Yes, the initial potential is 2500 V (with respect to the final one).
     
  13. Dec 10, 2014 #12
    :):w:):):):):):):):):w:w:w:w:D:D:D:D:D

    so if i did
    ½×M×Vo2 + q(Ua) = ½×M×Vf2 +q(Ub)
    0+(1.6x10-19c)(2500v)=½×M×Vf2 + q(0v)
    i would get 10/10?
     
  14. Dec 10, 2014 #13
    wait if i did this with a negative charge going towards the positive terminal against the field. and ΔV= 2500v
    i would get
    This again Vf = √(2(-1.6x10-2c)(2500V))
    the concept makes sense but math screws my concept up.
     
  15. Dec 10, 2014 #14

    ehild

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    10/10 what??? What is the mass of the isotope?
     
  16. Dec 10, 2014 #15
    0.666*10^-25kg but im just asking if what i did was right.

    and if i did it with a
    negative charge going towards the positive terminal against the field. and ΔV= 2500v
    i would get this again using conservation of energy eq KE + PE = KEf + PEf

    Vf = √(2(-1.6x10-2c)(2500V)/M)
    which wont work because of the negative.

    edit: sorry forgot about dividing by mass
     
  17. Dec 10, 2014 #16

    ehild

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    If a negative charge goes toward the positive terminal then its potential energy decreases, as PE=qU and q is negative.
    Initial position: UB=0, PEinitial=0, KE initial = 0.
    Final position: UA=2500 V, PEfinal =[ -1.6x10-19 C](2500 V), KEfinal+Pfinal=0.--> -1.6x10-19 *2500 +1/2 Mv2=0.
    Is there any problem?
     
  18. Dec 10, 2014 #17

    ehild

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    What you did with the positive charge was good. What you did with the negative charge, it was wrong.
     
  19. Dec 10, 2014 #18
    I love you.
     
  20. Dec 10, 2014 #19
    Wait Wait Wait IF the initial voltage is 2500v and final is zero wouldnt that make ΔV = -2500v
    instead of ΔV=+2500v like the question gave us.
    (when proton goes from A to B)
     
  21. Dec 10, 2014 #20

    ehild

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    Potential difference between points A and B is not the same as the change of potential along the route of a particle. The first refers too the set-up. The second refers to the moving particle.

    The potential difference of point A with respect to point B is 2500 V. The problem said that there is 2500 V potential difference between the plates, but it did no say in what sense. You had to find it out, from he sentence, "Motion of the positively-charged isotopes toward the right was initiated by a potential difference of 2500V". Potential difference meant the magnitude of the difference between the potentials here.
    If the positive isotope moved from the positive plate to the negative one, the potential changed from 2500 V to 0 V along its route, it travelled across -2500 potential difference. U final -U initial = -2500 V.
     
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