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Potential difference between plates

  1. Aug 27, 2008 #1
    a certain parrallel plate capacitator has a plate separation d and a potential difference between the plates of V. If the electric field between the plates is uniform
    a) What is the strength of the electric field F

    2. Relevant equations

    Coloumbs Law F=k(q1q2/d^2)
    Electric potential=electric potential energy/amount of charge

    3. The attempt at a solution

    i am not sure which equation to use here with the given variables i know that the potential difference is the change in velocity (i think thats right anyway)
  2. jcsd
  3. Aug 27, 2008 #2

    Doc Al

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    No, that's not right. Find the relationship between voltage, field, and distance: Work and Voltage: Constant Electric Field
  4. Aug 27, 2008 #3


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    you say capacitator, I say capacitahtor … !

    Hi annjolino! :smile:

    (if you'd spelt capacitor right, you'd have got the autolink to the PF Library … :rolleyes:)
    Nooo …

    Voltage = potential difference is energy/charge, so it's the same as work done per charge … and electric field is force/charge …

    and work done = force times … what? :smile:
  5. Aug 27, 2008 #4
    Hmmm bad day with spelling i guess....:rolleyes:

    work done = force x distance
  6. Aug 27, 2008 #5


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    … what we call … incapacitated! :biggrin:
    Yes … and so F = … ? :smile:
  7. Aug 27, 2008 #6
    does the F in this question equal the force of the electric field or does the F mean the E int the equation Electric field = force/charge.... i know that might osund a bit weird but the way i read the question F = E= f/c.....

    i think i am going the long way about this but so far i have arranged the three formulas to read this

    V=W/q which looks like V= (F/d)/(F/E)

    am i headed in the right direction???
  8. Aug 27, 2008 #7
    i just noticed a little hint in the question 'NB 1Vm-1 is equivalent to 1NC-1

    so from the Vm-1 does that mean that the F=V/d?????
    could it be that simple
  9. Aug 27, 2008 #8


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    Hi annjolino! :smile:

    Yes … the question is very clear …
    That means that F is the electric field. :smile:
    Not following that. :confused:

    (and what's E anyway? there's no E in the problem :rolleyes:)

    Write it out clearly … like this …

    V = Voltage = work done per charge

    Work = force x distance

    F = Electric field = force per charge

    So V/F = work / force = distance = d. :smile:
  10. Aug 27, 2008 #9
    part c of the question is, if the acceleration of a charged particle between the plates is a when it is halfway between the plates, why will the particle acceleration experience the same acceleration if it is only one quarter the distance between the plates? defend your answer...

    Ok so i think the the particle will experience the same acceleration at any point across the field due to the fact that the electric field between the plates is uniform... as stated by the question... i am just not sure how to defend my answer... i guess i have to use an equation of sorts??
  11. Aug 28, 2008 #10


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    You've already answered the question. Why do you need to defend it? If you want to put it more explicitly, you just need to show how force is related to E-field and acceleration of charged particle.
  12. Aug 28, 2008 #11


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    … you have a capacity for sagacity …

    Hi annjolino! :smile:

    Yes, that's fine … :smile:

    if you really want an equation, how about good ol' Netwon's second law … F = ma … combined with the equation for force from an electric field? :wink:

    ooh … EDIT:
    Hi Defennder! :smile:

    hmm … let's see … :wink:

    i] 'cos the question says so … :rolleyes:

    ii] against wild animals … :eek:

    iii] 'cos it can't defend itself … :blushing:

    iv] it's what Cicero would have done … :approve:

    v] look who's talking … Defennder :biggrin:
    Last edited: Aug 28, 2008

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