1. **Problem statement:** A current $i$ flows through a wire composed of copper (Cu) and aluminum (Al) sections joined end to end with identical cross-sections. We need to find the charge accumulated at the boundary between the two wires, given resistivity $\rho$, conductivity $\sigma$, and current density $J$. 2. **Understanding the physical setup:** The current $i$ is continuous through the wire, but due to different material properties (resistivity and conductivity), there can be a discontinuity in the electric field $E$ and potential $V$ at the boundary. 3. **Relation between current density and electric field:** $$J = \sigma E = \frac{i}{A}$$ where $A$ is the cross-sectional area (same for both wires). 4. **Electric field in each material:** $$E_{Cu} = \frac{J}{\sigma_{Cu}} = \frac{i}{A \sigma_{Cu}}$$ $$E_{Al} = \frac{J}{\sigma_{Al}} = \frac{i}{A \sigma_{Al}}$$ 5. **Potential difference across a small segment $\ell$ near the boundary:** $$V = E \ell$$ So the potential difference across the boundary is: $$\Delta V = E_{Al} \ell - E_{Cu} \ell = \ell \left( \frac{i}{A \sigma_{Al}} - \frac{i}{A \sigma_{Cu}} \right) = \frac{i \ell}{A} \left( \frac{1}{\sigma_{Al}} - \frac{1}{\sigma_{Cu}} \right)$$ 6. **Charge accumulation at the boundary:** The charge $Q$ accumulated at the boundary relates to the discontinuity in the electric displacement field $D = \epsilon_0 E$: $$Q = \epsilon_0 A (E_{Al} - E_{Cu}) = \epsilon_0 A \left( \frac{i}{A \sigma_{Al}} - \frac{i}{A \sigma_{Cu}} \right) = i \epsilon_0 \left( \frac{1}{\sigma_{Al}} - \frac{1}{\sigma_{Cu}} \right)$$ 7. **Expressing in terms of resistivity $\rho = \frac{1}{\sigma}$:** $$Q = i \epsilon_0 (\rho_{Al} - \rho_{Cu})$$ 8. **Comparing with given options:** - Option (1): $i \epsilon_0 (\rho_{Al} - \rho_{Cu})$ matches our derived expression. - Option (2): 0 (incorrect) - Option (3): $i \epsilon_0 (\sigma_{Al} - \sigma_{Cu})$ (incorrect, conductivity difference is not correct here) - Option (4): None **Final answer:** Option (1) $i \epsilon_0 (\rho_{Al} - \rho_{Cu})$