1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitance nd Spherical cells

  1. Nov 12, 2011 #1
    Hey all I'd like to help me..
    thnx in advance
    here z tha Q's :

    1) Two plane parallel conducting plates each have area S and are separated by a distance b.
    One carries a charge +Q ; the other carries a charge -Q. Neglect edge effects

    a) What is the charge per unit area on each plate ? where does it reside ?
    b) what is the electric field between the plates ?
    c) what is tha capacitance ?
    d) As the plate separation is increased, what happens to E, v and C ?
    e) If a dielectric is inserted b/w the plates, what happenes to E, v and C ?

    .........................
    2) The distance L that oxygen can diffuse in the steady state is approximately L= √(CD/Q)
    , where C is the oxygen concentration, D is the diffusion constant, and Q is the rate per unit volume that oxygen is used for metabolism

    a) show that L has dimensions of length
    b) The diffusion of oxygen in air is about 10,000 times larger than the diffusion of oxygen in water . By how much will the diffusion distance L change if oxygen diffuses through air instead of water, all other things being equal ?
     
    Last edited: Nov 12, 2011
  2. jcsd
  3. Nov 12, 2011 #2

    BruceW

    User Avatar
    Homework Helper

    You should try to solve the problem yourself first. Sorry, but physics forums has rules and I might get banned if I do the problem for you. For the first question, are you required to derive the result using Gaussian surfaces, or simply to use the equation for a capacitor? Either way, try to solve the question, and post your attempt.
     
  4. Nov 12, 2011 #3
    well okay I'll try .. but plz I need ur correction
    the 1st one bout simply eq. for capacitor
    ..
    so az I learned the first one its bout
    ( charge density ( sigma ) = charge (q) / area ( S ) ) and cus its bout one plate I think Ive to dividin the equ. by 2 !!!
    so it 'll be like
    Sigma = q/2*S
    reside ?? humm well in the middle between the 2 plates :\

    b)Electric field will be
    E= Sigma/ e_0 ; while sigma = Q/S
    so E= Q/S*e_0
    c) C= Q/V = Q/E*B = Q*S*e_0/Q*b = S*e_0/b
    but Ive got another relation which is C = kappa * e_0 * S / b
    which one should I take ?

    d)
    C will be decreased according to C = e *S/b
    and the E stay the same I think!!! and the potential will increased

    e) dielectric will increased tha C , and decreased the E .. and the potential I dunno :\
     
    Last edited: Nov 12, 2011
  5. Nov 12, 2011 #4
    plz guys any hint for the 2nd Q ??
     
  6. Nov 12, 2011 #5
    I'v doubts bout tha 2nd Q ..
    C = m-3
    D = m2/s
    Q= m-3/s
    so L = m ..
    is that right !!!
    while D grows by 10,000
    " assume the remaining is constant "
    L = 100
    is that tha better way to explain tha part B in Q 2 ?
     
  7. Nov 12, 2011 #6

    BruceW

    User Avatar
    Homework Helper

    a) The question says that the absolute value of the total charge per plate is Q, so you don't need to divide by 2. The question said the two plates were conducting, so for a conductor, where does the extra charge go? (I'm hoping you've been taught this).

    b)Yep, this is right.

    c)The first one is correct. I think the second one is using kappa as the relative permittivity, which would be used if there was a material between the plates.

    d)Yep, all correct.

    e)They are correct, but do you know why? Think of what the dielectric material does.
     
  8. Nov 12, 2011 #7
    a) the charge will be in the inside of each plates
    so I'm goin to take the first relation which is Sigma = Q/S
    am I right ?

    e) the dielectric materials prevent tha charges to flow through the materials so the Q will stay the same and doesnt make any change on the E
    right ?
     
  9. Nov 12, 2011 #8

    BruceW

    User Avatar
    Homework Helper

    You've done part a correctly. Part b is worded badly. It says "The diffusion of oxygen in air is about 10,000 times larger" And you've taken that to mean D is 10,000 times larger. I think you're probably right, but the question is vague.
     
  10. Nov 12, 2011 #9
    e) and the potential v will be decreased right ?
     
  11. Nov 12, 2011 #10
    so u think Ive done part B too ? any suggest ?
     
  12. Nov 12, 2011 #11

    BruceW

    User Avatar
    Homework Helper

    a) Sigma = Q/S is correct. And yes, the charge is confined to the plates, but think where the charges will want to go (it won't be in the middle of the plate).

    e)No, you got it right before when you said E will be less. Dielectric materials contain bound charges. you're right that there is no flow of charge, but the bound charges in the dielectric material have the ability to align in a certain way. There is an electric field going through the dielectric, so how do you think the bound charges will align themselves? And what effect does this have on the total electric field?
     
  13. Nov 12, 2011 #12

    BruceW

    User Avatar
    Homework Helper

    The question is worded badly, since 'diffusion is 10,000 times larger' is a very vague statement, and it could mean anything. But in the context of the question, it is most likely talking about D, so I think you did it right.
     
  14. Nov 12, 2011 #13
    maybe the charges will be distribute in each plates
    or maybe it will be going toward the +ve charge ..

    when the dielectric material placed the E its causing Dielectric polarization Because of dielectric polarization, positive charges are displaced toward the field and negative charges shift in the opposite direction

    but the potential when the dielectric material placed will be decreased because of wht ?
     
  15. Nov 12, 2011 #14

    BruceW

    User Avatar
    Homework Helper

    a)I'm not sure what you mean by 'it will go toward the +ve charge'... There are +ve charges on one plate and -ve charges on the other plate, so just use the rule that opposites attract, and where do the charges end up?

    e)Yes, the positive charges are displaced in the direction of the field and the negative charges in the opposite direction. So what effect does this have on the total electric field? And to find the potential, just use V=Eb
     
  16. Nov 12, 2011 #15
    will be goes toward the plate with -ve Q ..

    e) the total will be zero
     
  17. Nov 12, 2011 #16

    BruceW

    User Avatar
    Homework Helper

    I still don't understand 'goes toward the plate with -ve Q', sorry.
    e)Well, the bound charges will not be enough to totally cancel out the electric field, so the electric field will be less, but not zero.
     
  18. Nov 12, 2011 #17
    if the charge will reside inside each plates .. thats mean the E will be moving from the +ve charge plate to the -ve charge plate
    so there is no specific place for the charge !!!
    is it like this way ?
     
  19. Nov 12, 2011 #18

    BruceW

    User Avatar
    Homework Helper

    The charge is inside each plate yes. And yes, the E field goes from the +ve plate to the -ve plate. I'm not sure what you mean by no specific place for the charge..

    To try to explain it: lets say plate A has some extra positive charges and plate B has some extra negative charges. The charges in plate A will be attracted to the charges in plate B, so the charges in plate A move as close as they can to plate B, so they end up on the surface of plate A which is closest to plate B. And a similar thing will happen for the charges in plate B.
     
  20. Nov 12, 2011 #19
    well yeah I get it
    thank you thank you so much
    really appreciate it and thank you for your patient ..
    Really thank you
    my problems are solved ..
    thank you again
     
  21. Nov 12, 2011 #20

    BruceW

    User Avatar
    Homework Helper

    Cool. All's well that ends well.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Capacitance nd Spherical cells
Loading...