Alternative electrostatic potential

[andre220]

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.

 

[DEvens]

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.

 

[andre220]

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?