• # question_answer Two long straight cylindrical conductors with resistivities ${{\rho }_{1}}$ and ${{\rho }_{2}}$ respectively are joined together as shown in figure. If current I flows through the conductors, the magnitude of the total free charge at the interface of the two conductors is A) zero                               B) $\frac{({{\rho }_{1}}-{{\rho }_{2}})I{{\varepsilon }_{0}}}{2}$ C) ${{\varepsilon }_{0}}I|{{\rho }_{1}}-{{\rho }_{2}}|$    D)        ${{\varepsilon }_{0}}I|{{\rho }_{1}}+{{\rho }_{2}}|$

[c] Apply Gauss's law: $\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}~(outgoing\text{ }flux-incoming\text{ }flux)=\Delta (EA)$ $V=IR=\frac{I\rho \ell }{A}\Rightarrow \frac{V}{\ell }A=I\rho \Rightarrow EA=I\rho$ $\Rightarrow \,\,\,\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}=I|{{\rho }_{1}}-{{\rho }_{2}}|\,\,\,\,\Rightarrow \,\,\,\,{{q}_{in}}=I{{\varepsilon }_{0}}|{{\rho }_{1}}-{{\rho }_{2}}|$