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Electric potential inside non-conducting sphere

  1. May 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A nonconducting sphere has radius R = 3.71 cm and uniformly distributed charge q = +3.76 fC. Take the electric potential at the sphere's center to be V0 = 0. What is V in volts at radial distance (a) r = 1.00 cm and (b) r = R?


    2. Relevant equations

    I suppose:
    Electric field of spherical shell = ( 1 / (4 x pi x E) ) * ( q / r² )

    3. The attempt at a solution

    Completely stuck, don't even know where to begin > <.
     
  2. jcsd
  3. May 16, 2007 #2
    we have by gauss law (im using cgs): E=Q/r^2
    the direction is radial, so you have a differential equation of dV/dr=Q/r^2, you know to solve this don't you?
     
  4. May 7, 2010 #3
    I have a question: I have a nonconducting sphere with a 0.06m radius with a 2*10^-8 C charge inside. I need to find V at a 0.03m radius.

    From MathematicalPhycisist I have to find (using Gauss) how the electric field varies with distance. According to him its: E = Q/r^2. Im using the IS units i don't know how that changes Gauss' equation: E.da= dq/E0. So I replace E with its equivalent function: E= Q/r^2. in the differential equation: dV = E dr ---> dV = (Q/r^2) dr. I integrate to find the potential V and I get -Q(1/r) and I plug in 0.03m where the "r" is. I already have the answer to the problem and this way to solve the problem does not match with answer which is: 6.18 * 10^3 V. Can somebody please help me?
     
  5. May 7, 2010 #4
    I'm afraid that's a little unrelated, MathematicalPhysicist.
    That may be the form that Gauss' Law gives you for r>R, but inside the spherical shell, there is no charge, and therefore Gauss' Law gives you 0 field inside, and r <= R is what the question is asking.

    Since the difference in potential is the path integral of the E field, you find 0 difference in potential between any two points inside the spherical shell.

    The potential on the very edge of the shell is an interesting question though... My intuition tells me it's 0, but I'm finding it hard to convince myself that is the case...

    (Oh, the question may have been poorly phrased, with r standing for the radial distance from the surface of the shell, rather than from its center. If that is the case, then MathematicalPhysicist's advice is spot-on)
     
  6. May 8, 2010 #5
    I've been searching the internet for the solution and I've found a pretty good .pdf explaining how its supposed to be solved. Link:http://iweb.tntech.edu/murdock/books/v4chap4.pdf.

    Depending at where your reference point V=0 is, the way to solve this problem may change. If your reference point is at a distance infinity the solution is:

    (1/4*pi*Eo) * q(3R^2-r^2)/(2*R^2);

    Where the first part of the equation corresponds to the electric constant, k. q is the charge inside the non conductive sphere, R is the radius of the sphere (from the center to the surface), and r is the distance from the center to the point where you want to measure V.

    I haven't found the answer when its a shell. But in this its a solid sphere, and at the surface V is:

    (1/4*pi*Eo) * q(R^2);

    Well I hope this helps.
     
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