# Is the $E$ field near a thin cylindrical line charge stronger than a thick one at the same voltage?

Compare two infinite cylindrical conductors held at the same voltage $$V$$, with diameters $$R_1$$ and $$R_2$$. The electric field strength and voltage outside the conductor are $$E = \frac{\lambda}{2\pi\epsilon_0}\frac{1}{r},\space V = \frac{\lambda}{2\pi\epsilon_0}\ln{\left(\frac{R}{r}\right)}$$ where $$\lambda$$ is the linear charge density.

Naively one could expect that the thinner conductor is analogous to a sharp point on a charged conductor, and that the electric field near it is stronger. But is this actually the case? It's obvious that for constant charge density, you have a thinner Gaussian cylinder enclosing the same charge and so the field just goes up as $$1/r$$, but this is not obvious to me if only the voltage is held constant.

Instead of thinking of just a single conductor (which is difficult because the potential doesn't go to zero at infinity), consider one near a conducting grounded plane such that you can use an image line charge of the opposite sign:

(note: I had this problem while figuring out voltage ratings for insulated wires, which is what the purple circle and thickness are about, they're mostly irrelevant here)

Call the distance from the centre of the positive wire to the plane $$a$$. Since for a single wire $$V = \frac{-\lambda}{2\pi\epsilon_0}\ln\left(r\right)$$, the potential of the two wires is given by

$$V = -k\left[ \ln(|\mathbf{r}-\mathbf{r_+}|) - \ln(|\mathbf{r}-\mathbf{r_-}|) \right]$$

where $$k = \frac{\lambda}{2\pi\epsilon_0}$$, and $$|\mathbf{r}-\mathbf{r_+}|$$ is the distance from a point $$\mathbf{r}$$ to the positive wire. Now if you set the voltage at the surface of the wire to some $$V_0$$ and evaluate that at the nearest point of the wire to the plane, you find that

$$k = \frac{V_0}{\ln(2a-R) - \ln(R)}$$

Then from superposition of the two wires, each contributing $$E = k/r$$, the field strength at a point along the x-axis between $$x = -d$$ and $$x = d$$ is:

$$E = k\left(\frac{1}{|x+a|} + \frac{1}{|x-a|}\right)$$

Which you can then use to convince yourself, either by looking at the scaling or by plugging in values, that the field at the surface of a thinner conductor is larger at the same voltage (though it is a small effect for all but the tiniest wires).