Connecting wires end-to-end

alt text
At Steady State,
The Current Density ( $\displaystyle J$ ) is same for both the conductors.

Let $\displaystyle Q$ be the charge accumulated at the boundary between the two conductors.
Let $\displaystyle \sigma_1,\sigma_2$ represent the respective conductivities of the conductors.

Note that $\displaystyle J = \sigma E$ , where $\displaystyle E$ is the Electric Field inside the conductor.

Also note that,

$\displaystyle \sigma = \frac{1}{\rho}$ , where $\displaystyle \rho$ is the resistivity of the conductor.

Thus,
$\displaystyle J_1 = J_2$
$\displaystyle \sigma_1E_1 = \sigma_2E_2$

$\displaystyle \boxed{\Rightarrow E_2 = \frac{\sigma_1}{\sigma_2} E_1}$

MISTAKE IN FIGURE: Since the current is flowing in the same direction, so, the Electric Field must also be in the same direction (actually, I did it as a mistake in the figure, later realizing it).

Now, consider a Gaussian Surface , which is a cylinder of radius equal to that of the bigger conductor.
It is clear from applying Gauss' Law that,

$\displaystyle E_2A_1 - E_1A_2 = \frac{Q}{\epsilon_o}$

Substituting $\displaystyle E_2$ in terms of $\displaystyle E_1$ , we get,

$\displaystyle Q = A\epsilon_o\sigma_1E_1\left(\frac{1}{\sigma_2}-\frac{1}{\sigma_1}\right)$

We know, $\displaystyle I = JA = (\sigma E)A$ , therefore,

$\displaystyle Q = I\epsilon_o\left(\frac{1}{\sigma_2}-\frac{1}{\sigma_1}\right) = I\epsilon_o(\rho_2-\rho_1) = I\epsilon_o(\rho_{Al} - \rho_{Cu})$

$\displaystyle Q \approx \boxed{9.7\times 10^{-18}C}$