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!

Electric potential.

  1. Oct 15, 2011 #1
    Excuse me, may i know how should one calculate the electric potential of two conductors placed together? one inner sphere(radius a) together with another outer shell(radius b). Why is the "electric potential between the sphere and shell" equal to the electric potential in the sphere? LEts put the charge of inner one be q1 and outer one be q2. How we find the potential everywhere? For a>r, a<r<b, and also b<r. I understand about finding the outermost potential only.
     
    Last edited: Oct 15, 2011
  2. jcsd
  3. Oct 15, 2011 #2
    Use gauss's law to find the formula for the electric field. E(a<r<b)=Q/4 pi r^2
    Integrate this with your limits to find V(r) = Q/4 pi r
     
  4. Oct 15, 2011 #3
    But, the answer given is quite complicated. How do we find electric potential for two conductors actually? The inner one is affected by outer one or outer one is affected by inner one or both are unaffectable by any one? Let's say this case, the potential for r > b is easy , that is find the total net q and then the electric field and then the potential through integration. But for a<r<b, i have no idea. It depend on outer b or inner a? and also for a>r. Seriously, what to follow?

    something like this:http://webphysics.davidson.edu/physlet_resources/bu_semester2/c06_spheres.html" [Broken]

    but i do not understand well..

    Code (Text):
    For R < r < 1.5R, V(r) = kQ/r - kQ/1.5R - 2kQ/R
    From the previous region, we know that V(1.5R) = -2kQ/R. We also know how the potential changes as we move closer to a point charge:

    VA - VB = kq/rA - kq/rB

    Here VA = V(r), VB = V(1.5R) = -2kQ/R, q=+Q, rA = r, and rB = 1.5R.

    For r < R, V(r) = [-kQ/6R](7+3r2/R2)

    From the previous region, we have V(R) = -5kQ/3R.

    We derived previously that the potential difference between a point on the edge of an insulating sphere and a point inside is:

    V(R) - V(r) = -[kQ/2R3] (R2-r2)

    Solving for V(r) gives:

    V(r) = -5kQ/3R + kQ/2R - kQr2/2R3

    V(r) = -10kQ/6R + 3kQ/6R - 3kQr2/6R3

    For r < R, V(r) = [-kQ/6R](7+3r2/R2)
    starting from this part , i start to blur..
     
    Last edited by a moderator: May 5, 2017
  5. Oct 15, 2011 #4
    In conductor, all the charge "flows" (following the divergence lines) to the surface. That means there is no charge under the surface of the conductor (the equivalent of saying that the electric field inside a conductor is zero).

    (1) The inner sphere has a charge of q1. This means that if the inner sphere is placed inside an uncharged conducting shell, the outer surface of the uncharged conducting shell must have an induced charge of q1 on its surface (in order for Coulomb's law to hold).

    However, the uncharged conducting shell didn't have a charge to begin with, so there must be a charge of -q1 distributed in some way in the conducting shell for the net charge of the outer shell to be zero. However, a charge cannot exist inside a conductor (since the charge flows following the electric field lines), so this -q1 must be on the inner surface of the uncharged conducting shell.

    (2) A shell has a charge of q2. This means that the outer surface of the shell contains a charge of q2.

    The original configuration of the problem (the sphere of charge q1 inside a shell of charge q2) is merely a superposition of (1) and (2).

    To find the electric field, you'd have to work from inside out. The reason is that if you work from outside in, when you reach the innermost sphere, there would be no way to induce a charge because a solid sphere only has one surface, whereas a shell has two surfaces.

    Also, the shell theorem states that the field inside a spherical shell with constant charge density is zero, and of course, a sphere is a superposition of multiple shells, so you can just ignore the outer layers when you calculate for the electric fields.
     
  6. Oct 15, 2011 #5
    Ok, Thanks for your respond. But,how to get the electric potential for the spheres? I mean any specific terms or formulas we have to use?
     
  7. Oct 15, 2011 #6
    Integrate the electric field equation with respect to r and multiply by -1.
     
  8. Oct 15, 2011 #7
    Ok, but for concentric spheres? Should or can we use the same thing? The answer comes out to be very different with the one my lecturer gives me.
     
  9. Oct 17, 2011 #8
    inside the first sphere the potential is 0 or whatever it is to fit the boundary conditions because any 1/r^2 field is 0 inside a uniform spherical shell

    outside the second it's just q1+q2/r^2
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electric potential.
  1. Electric potential (Replies: 1)

  2. Electric potential (Replies: 3)

  3. Electric Potential (Replies: 2)

Loading...