Alternative electrostatic potential

  • Thread starter andre220
  • Start date
  • #1
75
1

Homework Statement


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##.

Homework Equations


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

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.
 

Answers and Replies

  • #2
DEvens
Education Advisor
Gold Member
1,203
457
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.
 
  • #3
75
1
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?
 

Related Threads on Alternative electrostatic potential

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
976
  • Last Post
Replies
2
Views
8K
Replies
1
Views
384
Replies
1
Views
2K
Replies
5
Views
4K
Replies
2
Views
924
Top