1. Not finding help here? Sign up for a free 30min 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!

Equipotential Spheres

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A spherical potential conductor of radius R1 is charged to 20kV. When it is connected by a long, fine wire to a second sphere faraway. its potential drops to 12kV. What is the radius of the second sphere.


    2. Relevant equations

    V=kq/r


    3. The attempt at a solution

    So I know that R2= R1* (q2/q1). Thus, I need to find the ratio q2/q1. I am kinda lost in what to use to figure out q2/q1 ratio.

    k*q1/R1 = 2x10^4 and k*q2/R2= 4x10^3 (because 2 spheres are equipotential at 12kV and sphere 1 drops 8kV so sphere 2 has to gain 8kV so initially sphere 2 is at 4kV)

    k*q1/R1 = k*q2/R2 = 12x10^3.

    With these 3 equations, I am still unable to figure out what is q2/q1. Can you guys give me some hints and suggestions ?
     
  2. jcsd
  3. Oct 24, 2009 #2
    Sphere 2 only will gain the same amount of voltage if it has the same radius.
    I think you are supposed to assume sphere 2 is initially uncharged. If it has an initial charge, you can't do the problem. Sphere 2 could be gigantic, with a potential slightly below 12 kv, or tiny with a very large negative potential.

    the spheres act as capacitors with capacity r/k
     
  4. Oct 24, 2009 #3
    There is no need to consider them as capacitors.
    Your equations are almost there, except for the part on sphere 2 being at 4kV initially of course, but do take care of the variables (why do i see q1 appear twice in different situations?)
    Hint: Charge is conserved throughout the transfer process
    That should allow you to simplify the expressions.
     
  5. Oct 24, 2009 #4
    So if I assume that sphere 2 is uncharged initially, then I should make its potential equal to 12kV right ? since 2 spheres are equipotential.

    Oh, if charged is conserve then kq1/R1= 20kV = kq1/R1= 12kV ?
     
  6. Oct 25, 2009 #5
    Yes. Sphere 1 is at 20kV and Sphere 2 at 0V initially, with both becoming 12kV after charge flows between the two spheres

    Er..no...that equation clearly doesn't make sense - 20 = 12??
    Charge conservation in the sense that if Sphere 1 possesses Q coulombs of charge initially, and q2 coulombs of charge flow from Sphere 1 to 2, leaving sphere 1 with q1 coulombs remaining, then clearly q1 + q2 = Q.
    Then your originally formulated equations can be written as:
    k*(q1+q2)/R1 = 2x10^4
    k*q1/R1 = k*q2/R2 = 12x10^3
    I guess you can carry on?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equipotential Spheres
  1. Equipotential surfaces (Replies: 0)

  2. Equipotential Contours (Replies: 8)

  3. Equipotential Surfaces (Replies: 1)

  4. Equipotential Surfaces (Replies: 2)

Loading...