Question

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. $\frac{({\rho }_1-{\rho }_2)I{\epsilon }_0}{2}$
• C. ${\epsilon }_{0}I|{\rho }_1-\rho |$
• D. ${\epsilon }_{0}I|{\rho }_1+\rho |$

Answer 1

Answer: (C)

${\epsilon }_{0}I|{\rho }_1-\rho |$

Apply Gauss's law : $\frac{q_{in}}{{\epsilon }_0}=$ (outgoing flux - incoming flux) $=\Delta \left(EA\right)$
$V=IR=\frac{I\rho l}{A}\Rightarrow \frac{V}{l}A=I\rho \Rightarrow EA=I\rho$
$\Rightarrow \frac{q_{in}}{{\epsilon }_0}=I|{\rho }_1-{\rho }_2|\Rightarrow q_{in}=I{\epsilon }_0|{\rho }_1-{\rho }_2|$