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Find charge and radius of conducting spheres given voltage and electr

  1. Feb 12, 2014 #1
    The electric potential immediately outside a charged conducting sphere is 200 V, and 10.0 cm farther from the center of the sphere the potential is 150 V. Determine (a) the radius of the sphere and (b) the charge on it. The electric potential immediately outside another charged conducting sphere is 210 V, and 10.0 cm farther from the center the magnitude of the electric field is 400 V/m. Determine (c) the radius of the sphere and (d) its charge on it. (e) Are the answers to parts (c) and (d) unique?


    for the first part I said 200=kq/r and 150 =kq/(r+0.1) to solve for r=0.3m I then input this in and solved for q=6.67x10^-9C
    For the second part I said 210=kq/r and 400=q/(4pi(r+0.1)^2) and get r=-0.1 which cannot be right! help would be appreciated thanks
     
  2. jcsd
  3. Feb 12, 2014 #2

    Simon Bridge

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    Hint: r appears squared - so you should have two solutions.
     
  4. Feb 12, 2014 #3
    I know there should be two but with 1600pi r^2+(320pi-210/k)r +16pi I only get r=-0.1
     
  5. Feb 13, 2014 #4

    ehild

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    The equation in red should be 400=kq(r+0.1)2. (K=1/(4piε0), you missed ε0)

    ehild
     
  6. Feb 13, 2014 #5

    Simon Bridge

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    You are saying that from: 1600pi r^2+(320pi-210/k)r +16pi = 0
    you only get r=-0.1 ?

    But you can't...
    The discriminant is (320pi-120/k)^2-4(1600pi)(16pi) < 0 ... so you should have no roots at all.

    This means you set up the equation wrong.
    ehild may have spotted the answer - you may have made other errors: recheck your working.

    I'd start from scratch:
    So you have charge q giving voltage V at radius R, and electric field E at radius R+r.
    You must find q and R given V, E, and r. Leave the numbers to after doing the algebra.

    shortcut: Write the equations in the form kq=...
     
    Last edited: Feb 13, 2014
  7. Feb 13, 2014 #6
    ok thanks so much I had forgot the e in the equation
    so setting equal for kq I get 210r=400(r+0.1)^2 but then I still get negative square roots...?
     
  8. Feb 13, 2014 #7

    Simon Bridge

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    Please show your working - the trouble is either algebra or arithmetic.
    Please follow advise and wait till after the algebra is done to plug the numbers in.

    Start with your ##Vr=E(r+d)^2## - divide through by E, put in standard form and use the quadratic equation.

    Then put in: V=210V, E=400V/m, d=0.1m
     
    Last edited: Feb 13, 2014
  9. Feb 13, 2014 #8
    ok so 210=kq/r and 400=kq/(r+0.1)^2 setting equal gives 210r=400(r+0.1)^2 which gives the quadratic 400r^2+80r+4=210r which gives 400r^2-130r+4 ....nevermind I had put -40r for some reason! thanks so much
     
  10. Feb 13, 2014 #9

    Simon Bridge

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    This is why I kept telling you to put the numbers in last.
    If you do all your algebra with the variables this sort of mistake is much less common and much easier to spot when it does happen.

    I realize that many students are uncomfortable with this approach but it is worth it in the long run: not all your problems will be this simple.
     
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