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Electric potential probelm HELP PLEASE TEST TMRW!

  1. Apr 5, 2013 #1
    Electric potential probelm!! HELP PLEASE TEST TMRW!

    Two uniformly charged spherical shells of radii R and 2R with identical surface charge densities σ are along the x-axis. The surface of each of the spheres is located a distance R from the origin O
    Define the potential at ∞ to be zero.

    Along the x-axis are two locations A and B that are R/2 from the Origin (O).

    A) What is the potential at A?
    B) What is the potential at B?
    C) Work required by external foce to move a charge q from A to B?
    D) Where between the spheres and along x-axis is the potential the minimum?

    Can someone explain this problem to be conceptually on what is going on because i dont uderstand.

    this is what i have done so far
    A) V = Kq / R ----> q = σ x 4πr2 and distance r = R/2
    so VA = 2K(σ x 4π(R)2) / R

    B) is the potential at B the same as the potential at A except for the radius is 2R instead of R?
    VB = 2K(σ x 4π(2R)2) / R

    c)W = qΔV
    ΔV = answer to B - answer to A = VA = 2K(σ x 4π(3R)2) / R ???

    D)what does this mean??? potential the minimum???!

    also my teacher did not provide me with a picture or any diagram so i'm kinda visualizing it in my head. HELP ME PLEASE!
     
  2. jcsd
  3. Apr 5, 2013 #2
    Check out the attachment. Maybe this is what the diagram may look like.
     

    Attached Files:

  4. Apr 5, 2013 #3

    THANKS!

    So would this be correct
    Vsphere on the left = Kq / R ----> q = σ x 4πr2 and distance r = 3R - R
    so VA = K(σ x 4π(2R)2) / 2R

    Vsphere on the right = K(σ x 4π(2R)2) / 4R
    K(σ x 4π(R)2) / 4R

    im confused on the distance part. is Sphere on the left 2R away from point A and right sphere is 4R away from point A?
     
  5. Apr 5, 2013 #4
    also im still confused on the part d where it says where the potential is at minimum. how would i solve that part?
     
  6. Apr 5, 2013 #5
    Sorry for the late reply.

    In the diagram, A is R/2 from the origin. So its distance from the center of the 2R sphere is 5r/2. And its distance from the R sphere is again 5R/2. So you have to find VA as the sum of the two potentials because of the two spheres.

    To find the point where the potential is minimum, consider a point (x,0). Find its potential. Differentiate the expression w.r.t. x and equate to zero. You'll get a value of x where the potential is minimum.
     
  7. Apr 5, 2013 #6

    wait hows it 5R from the origin? The center of the 2R sphere is 3R from the origin? and the center of R sphere is 2R from the origin?
     
  8. Apr 5, 2013 #7
    How's what 5R from the origin?
     
  9. Apr 5, 2013 #8

    You said the distance from the center of the 2R sphere is 5r/2. And its distance from the R sphere is again 5R/2. i dont understand how its 5R from the point of the sphere to the origin (0,0) on the graph
     
  10. Apr 5, 2013 #9
    also what does the expression w.r.t. mean
     
  11. Apr 5, 2013 #10
    I'm really sorry for my poor use of grammar there. Let me explain again. From the center of the sphere of radius 2R, the distance of A is 5R/2 right? Also from the sphere of radius R, the distance of A is 5R/2. So now find the potential at A because of the two spheres.

    w.r.t means with respect to. Differentiate the expression for potential at (x,0) w.r.t. x and equate it to zero to obtain the x where the potential in minimal.
     
  12. Apr 5, 2013 #11
    i see. so for point b, the two spheres would have different distances right?
     
  13. Apr 5, 2013 #12
    Yes. Point B would have different set of distances from the centers of the spheres. Find the potential there too.

    I hope you have understood how to obtain the point where the potential is minimum.
     
  14. Apr 5, 2013 #13

    i dont really understand the minimum potential part.
     
  15. Apr 5, 2013 #14
    Ok. First choose an arbitrary point (x.0). What is the net potential at this point?
     
  16. Apr 5, 2013 #15
    as in we can choose any x value? we equate the potential to zero?
     
  17. Apr 5, 2013 #16
    Yes any value of x. Don't equate the potential to zero. Just write the expression first.

    Edit: Write the potential in terms of the variable x
     
  18. Apr 5, 2013 #17
    v = kq/x
     
  19. Apr 5, 2013 #18
    No! No!

    What is the potential at (x,0) due to the two spheres? How did you find potential at A? Do the same thing but choose the point(x,0)
     
  20. Apr 5, 2013 #19
    the potential due to the two spheres at point A is the sum of the potential of the two sphere so kq/r + kq/r with r = the distance right?

    so since we're using the point x,0 then shouldn't it be kq / x?
     
  21. Apr 5, 2013 #20
    You've got the concept but the equation is still a bit wrong. Firstly the charges on both the spheres is not the same right? Then the distance of the point (x,0) from both the spheres in not the same, is it?
     
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