Voltage of a point in varying permittivity medium problem

[null void]

Homework Statement

- A point charge is placed at the origin of the medium.
- The relative permittivity of the medium, \varepsilon_r = a / r, a is a constant, r is the radius from origin to any point around the charge.
- Objective of this question is to find the expression for voltage.at any point around Q

Homework Equations

The Attempt at a Solution

\vec E = \frac{Q}{4\pi \varepsilon_0 \varepsilon_r r^2} \vec r
= \frac{Q}{4\pi \varepsilon_0 (\frac{a}{r}) r^2} \vec r
= \frac{Q}{4\pi \varepsilon_0 a r }\vec r

V = - \int_\infty^r{\vec E \cdot d \vec l}
= - \int_\infty^r{ \frac{Q}{4\pi \varepsilon_0 a r } dr}
= - \frac{Q}{4\pi \varepsilon_0 a}[ln r - ln \infty]
= \frac{Q}{4\pi \varepsilon_0 a}[\ln \frac{\infty}{r}]

In the integration i use \infty as lower boundary to assume the infinity is at 0 voltage.
But the expression seems to be wrong, anyone know where did I messed up?

 

[theodoros.mihos]

Try to work by Gauss's Law in differential mode.

 

[null void]

Try to work by Gauss's Law in differential mode.

Sorry, I don't quite get you. I have searched around about that differential form, I am not sure what should i find with it.
Q = \int\int D \cdot ds = \int\int\int \rho_v dv
The Q is given in the question as a constant, D can be derived out by Coulomb's law.

The answer for the question is
V = \frac{Q}{4\pi \epsilon_0 a}\ln{r}

it seems as if the integration(in my 1st post) takes the lower boundary at r=1 and upper boundary at r = r

 

[theodoros.mihos]

I think than there is a trick on integration.

In the integration i use ∞\infty as lower boundary to assume the infinity is at 0 voltage.

Integration limits driven by $dl$ incrase. So I think the correct integral is:
$$ V = -\int_r^\infty \mathbf{E}\cdot{d\mathbf{l}} \Rightarrow V = \frac{Q}{4\pi\epsilon_0a}\,\ln{r} - V_\infty $$
and now you can take $V_\infty=0$.

 

[TSny]

\vec E = \frac{Q}{4\pi \varepsilon_0 \varepsilon_r r^2} \vec r
= \frac{Q}{4\pi \varepsilon_0 (\frac{a}{r}) r^2} \vec r
= \frac{Q}{4\pi \varepsilon_0 a r }\vec r

V = - \int_\infty^r{\vec E \cdot d \vec l}
= - \int_\infty^r{ \frac{Q}{4\pi \varepsilon_0 a r } dr}
= - \frac{Q}{4\pi \varepsilon_0 a}[ln r - ln \infty]
= \frac{Q}{4\pi \varepsilon_0 a}[\ln \frac{\infty}{r}]

In the integration i use \infty as lower boundary to assume the infinity is at 0 voltage.
But the expression seems to be wrong, anyone know where did I messed up?

Your work looks correct. As you've discovered, you run into a problem if you try to set the potential equal to zero at infinity. The same thing happens for the potential due to an infinitely long, uniformly charged line of charge.

So, you can just pick an arbitrary point for V = 0. As you stated in post #3, it looks like the given answer takes V = 0 at r = 1. However, the overall sign is incorrect in their answer. V should decrease as r increases, if Q is positive.

 

[theodoros.mihos]

V should decrease as r increases, if Q is positive.

That is correct for constant $\epsilon$. The same calculation in this case produce the known result.
Behaviour is the same as a charge $Q$ leaves on a 2D world.

 

[TSny]

That is correct for constant $\epsilon$. The same calculation in this case produce the known result.
Behaviour is the same as a charge $Q$ leaves on a 2D world.

We haver \vec E = \frac{Q}{4\pi \varepsilon_0 \varepsilon_r r^2} \hat r which is valid for any "Class A" dielectric, even when $\varepsilon_r$ is not constant.

If Q is positive you can see that $\vec E$ will be radially outward from Q as long as $\varepsilon_r$ is positive (as usual). In this case, V must decrease as r increases.