You can think about it dynamically. I’ll restrict to 1D, so $\rho_1,\rho_2$ are linear resistivities (unit $\Omega/m$ in SI). Your two resistances have respective lengths $L_1+L_2=L$, resistances $R_1=L_1\rho_1,R_1=L_1\rho_1$ with currents $I_1,I_2$ and voltages $V_1+V_2=V$.
Say you impose a voltage $V$ and start with no charges accumulated at the junction. This means that the electric field is the same so:
$$
E=\frac{V_1}{L_1}=\frac{V_2}{L_2}
$$
Which gives:
$$
\begin{align}
V_1&=\frac{L_1}{L}V & V_2&=\frac{L_2}{L}V \\
I_1&=\frac{V}{L\rho_1} & I_2&=\frac{V}{L\rho_2}
\end{align}
$$
In particular $I_1\neq I_2$. To preserve conservation of charge, you deduce that you have an accumulation of charge $Q$ at the junction:
$$
\dot Q=I_1-I_2
$$
Accounting in general for the excess charge, using Gauss’s law and assuming you are in the quasi static regime, the original equation becomes:
$$
-\frac{V_1}{L_1}+\frac{V_2}{L_2}=\frac{Q}{\epsilon_0}
$$
which gives:
$$
\begin{align}
V_1 &= \frac{L_1}{L}V-\frac{L_1L_2}{2\epsilon_0L}Q & V_2 &= \frac{L_2}{L}V+\frac{L_1L_2}{2\epsilon_0L}Q \\
I_1 &= \frac{V}{L\rho_1}-\frac{L_2Q}{2\epsilon_0\rho_1L} & I_2 &= \frac{V}{L\rho_2}+\frac{L_1Q}{2\epsilon_0\rho_2 L}
\end{align}
$$
This results in the ODE:
$$
\dot Q +\left(\frac{L_2}{\rho_1}+\frac{L_1}{\rho_2}\right)\frac{Q}{2\epsilon_0L}=\left(\frac{1}{\rho_1}-\frac{1}{\rho_2}\right)\frac{V}{L}
$$
You therefore obtain and exponential relaxation to the equilibrium value:
$$
Q_\infty=2\epsilon_0\frac{\rho_2-\rho_1}{R_1+R_2}V
$$
Intuitively, in the stationary regime, the excess charge enhances the field in the region of high resistivity. Conversely, it suppresses the field in low resistance. This balances out the resulting currents stopping the charge pileup.
Hope this helps.
