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Electric field strength for a radial field.

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Q. A metal sphere of radius 20 cm carries a positive charge of 2.0 μC.

    a. What is the electric field strength at a distance of 25 cm from the centre of the sphere?

    b. An identical metal sphere carrying a negative charge of 1.0 μC is placed next to the first sphere. There is a gap of 10 cm between them. Calculate the electric force that each sphere exerts on the other.

    c. Determine the electric field strength midway along a line joining the centres of the spheres.


    2. The attempt at a solution

    I've got 2.88x105 Vm-1 for (a).

    For (b), I've got -0.072 N, but the book gives the answer as a positive value, whereas the product of the charges gets you a negative value. I don't get why it's negative. That's one problem.

    The real problem I suppose I've got is at (c). I've been at it for two hours now, and I don't get the answer the book has. The book says it's 4.72 Vm-1 but I have absolutely no idea how to get that. What freaking formula am I supposed to use to get that? I mean I've got two charges, but the single formula I've got needs one value for charge. What value do I use?! Please help!
     
    Last edited: Feb 4, 2012
  2. jcsd
  3. Feb 4, 2012 #2

    tiny-tim

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    hi gabloammar! :smile:
    i don't know the convention, but i think the idea is that the force is attractive, so is positive :confused:
    electric field is a vector, and so obeys the law of vector addition (ie, two or more add like vectors)

    so just find the two individual fields, and add (as vectors) :wink:
     
  4. Feb 4, 2012 #3
    The convention I've got in my book [or at least the one asked by my board] is that negative is attractive and positive is repulsive. But okay on that one, I'll sort it out in a little while.


    But for the other part, I don't know how to bring vectors into use in this case. I think I can only use equations. I'm not sure! :( Help!



    Edit: I don't get what you mean by just find the two electric fields. ? :(
     
  5. Feb 4, 2012 #4

    tiny-tim

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    hi gabloammar! :wink:
    each sphere has its own electric field (with the formula that you know), and you can add them

    (i think that's called the principle of superposition)

    remember, a field isn't just a number, it has a direction also (towards or away from the centre), and that makes it a vector! :smile:
     
  6. Feb 4, 2012 #5
    The formula that I know gives me the electric field strength for an electric field, so umm, is that what you're talking about? [sorry if I'm coming off as being weak at physics, this chapter's just really got me confused. Vectors + electricity = nightmare for me]

    [isn't the principle of superposition that thing where the total displacement of two interfering waves is equal to the sum of the displacements of the two waves?]

    and yeah that's what I read, that since force is a vector quantity, electric field strength is also a vector, but I'm not getting on how to apply that here because I've got no visual representation neither in the book [where the question is asked] nor in my mind :(

    I feel like an idiot but still.. Helpp! And thanks!
     
  7. Feb 4, 2012 #6

    tiny-tim

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    ok, draw one sphere red, and the other sphere blue

    at any particular point, draw the individual force vector from the red sphere as a red arrow (with the correct length, and pointing in the correct direction), and draw the force vector from the blue sphere as a blue arrow

    now add the arrows as vectors :smile:
     
  8. Feb 4, 2012 #7
    Okay you're practically teaching me something new here so bear with me please.

    I drew one sphere, wrote R for red inside it. Then I drew another one and wrote B for blue in it.

    After that you asked me to draw the individual vectors showing force, keeping in mind to make the lengths are accurate and the directions are correct.

    Directions: Assuming R is positive and B is negative, I draw the red line towards B and the blue line towards R, correct?

    Lengths: Here's where I'm puzzled. I've got values for the charges, +2 and -1 for R and B respectively. Where do I go from here? Or do I have to use any of the values from the previous two answers?
     
  9. Feb 4, 2012 #8

    tiny-tim

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    no, R repels (a positive test charge), so the red arrow point away from R :wink:
    yes, the length of each arrow is the field strength, the value you got before :smile:
     
  10. Feb 4, 2012 #9
    Did a little more research [mainly just read in some more resource books] and came back and read your bit and voila! Got the answer :) although one problem still!

    My answer is 4.32x105 while the book says 4.32 Vm-1. I don't get it :(
     
  11. Feb 4, 2012 #10

    tiny-tim

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    :confused: i think it must be a misprint

    does the book have 105 for answers (a) and (b) ?
     
  12. Feb 4, 2012 #11
    Yes for (a) but not for (b) :(
     
  13. Feb 4, 2012 #12

    BruceW

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    (b) should be of the order 10^6 smaller than (a), because when you calculate the force, you're multiplying by the charge, which is 10^-6 C. So (a) and (b) look fine.

    I'd agree with tiny-tim, it looks like their answer for c) is a misprint.
     
  14. Feb 4, 2012 #13

    tiny-tim

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    (oh yes, i forgot … (b) was a force :rolleyes:)

    the book's (a) and (c) can't possibly both be correct
    Thanks, BruceW! :smile:
     
  15. Feb 4, 2012 #14
    Yep I agree, it must be wrong. Thanks a lot for your help!
     
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