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!

Alternative electrostatic potential

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Assume that the electrostatic potential of a point charge ##Q## is $$ \Phi(r) = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^{1+\delta}},$$
    such that ##\delta \ll 1##.

    (a) Determine ##\Phi(r)## at any point inside and outside a spherical shell of radius ##R## with a uniform surface charge ##\sigma##.

    (b) If two concentric spherical conducting shells of radii ##a## and ##b## are connected by a thin wire, a charge ##q_a## resides on the outer shell and charge ##q_b## resides on the inner shell. Determine the ratio of charges ##\frac{q_a}{q_b}## to the first order in ##\delta##.

    2. Relevant equations
    $$E=-\vec{\nabla}\Phi$$
    $$Q = \sigma A = 4\pi R^2\sigma$$
    $$V = -\int \vec{E}\cdot \vec{dl}$$

    3. The attempt at a solution
    In the case when there is no ##\delta##: $$V(r>R) = -\int\limits_\infty^r\frac{1}{4\pi\epsilon}\frac{Q}{r^2}dr = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}$$
    $$V(r<R) = -\int\limits_\infty^R\frac{1}{4\pi\epsilon}\frac{Q}{r^2}dr = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$

    But...
    I have no idea what to do here, since if we were given the equation for ##E## I think it would make more sense.

    Any help is appreciated.
     
  2. jcsd
  3. Feb 10, 2015 #2

    DEvens

    User Avatar
    Education Advisor
    Gold Member

    You are given phi for one charge. When you have more than one charge how do you get the combined phi? What do you do when you have a charge distribution?

    You can check that you have done this correctly by checking that you get the correct V for the case of delta = 0. But you should probably do it for the general case and take the limit of delta going to zero rather than starting with delta = 0.
     
  4. Feb 10, 2015 #3
    In the case of a charge distribution I would integrate. So then I would just evaluate: $$ \Phi = \frac{1}{4\pi\epsilon_0}\int \frac{dQ}{r^{1+\delta}} $$ Inside we would have: $$\Phi(r<R) = \frac{1}{4\pi\epsilon_0}\int\limits_0^r \frac{dQ}{r^{1+\delta}}$$ and outside
    $$\Phi(r>R) = \frac{1}{4\pi\epsilon_0}\int\limits_r^\infty \frac{dQ}{r^{1+\delta}}$$
    Right?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Alternative electrostatic potential
Loading...