From <PERSON>'s law, at steady state, $$ \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} = 0 $$ so since $\vec{E}$ is curl-free, it is the (negative) gradient of the electrostatic potential $V$: $$ \vec{E} = -\vec{\nabla}V. $$
From the continuity equation and <PERSON>'s law, at steady state and for uniform electrical conductivity $\sigma = 1/\lambda$, $$ \vec{\nabla} · (\sigma \vec{E}) + \frac{\partial\rho}{\partial t} =0 $$ $$ \vec{\nabla}\sigma · \vec{E} + \sigma\vec{\nabla}·\vec{E} + \frac{\partial\rho}{\partial t} = 0 $$ $$ \vec{\nabla} ·\vec{E} = 0 $$ $$ \vec{\nabla}^2 V = 0. $$ This is the differential equation you must solve (<PERSON>'s equation), with constant potential boundary conditions on the spherical boundries.
I'm not sure if an analytical solution exists (I wouldn't be surprised if it does), but we can obtain approximate solution for $V$ for small $a$. If there were two point charges $Q$ and $-Q$ at $\vec{d}$ and $-\vec{d}$ in vacuum, the potential at points not on these charges also satisfies <PERSON>'s equation. In this simplified problem, according to <PERSON>'s law, the potential in the vicinity of the point charge at $\pm\vec{d}$ would be $$ V(\vec{r}) \approx \pm \frac{A}{\lVert\vec{r}\mp\vec{d}\rVert} $$ where $A$ is some proportionality constant that is yet to be determined. If $ a \ll d $, a distance $a$ away from $\pm\vec{r}$, we enforce
$$ V = \pm \mathcal{E} \approx \pm \frac{A}{a} $$ $$ A = a\mathcal{E} $$
The potential everywhere is given approximately (to the extent $a \ll d$ is valid) by $$ V(\vec{r}) = a\mathcal{E}\left(-\frac{1}{\lVert\vec{r} + \vec{d}\rVert}+\frac{1}{\lVert\vec{r} - \vec{d}\rVert}\right). 649
$$ You can see here why $a = 0$ is problematic. Then, $$ \vec{E} = -\vec{\nabla}V = a\mathcal{E}\left(-\frac{\vec{r} + \vec{d}}{\lVert\vec{r} + \vec{d}\rVert^3}+\frac{\vec{r} - \vec{d}}{\lVert\vec{r} - \vec{d}\rVert^3}\right) $$ and $$\vec{J} = \sigma \vec{E} = \sigma a\mathcal{E}\left(-\frac{\vec{r} + \vec{d}}{\lVert\vec{r} + \vec{d}\rVert^3}+\frac{\vec{r} - \vec{d}}{\lVert\vec{r} - \vec{d}\rVert^3}\right). $$ The charge density can be determined from <PERSON>' law: $$ \rho = \vec{\nabla}·(\epsilon \vec{E}) = \vec{\nabla}\epsilon · \vec{E} + \epsilon \vec{\nabla}·\vec{E} = \vec{\nabla}\epsilon · \vec{E}. $$ Given $\epsilon$ as a function of position, you can use this to find $\rho$. If $\epsilon$ is uniform, then we simply have $\rho = 0$. This is typically a good approximation within a conductor at steady state. 418