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Two cylindrical conductors with equal cross sections and different resistivity $ρ_{1}$ and $ρ_{2}$ are put end to end. Find the charge at the boundary of the conductors if a current $I$ flows from 1 to 2.

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$\frac{I ε_{0}}{ρ_{1} - ρ_{2}}$

$\frac{I ε_{0}}{ρ_{2} - ρ_{1}}$

$(ρ_{1} - ρ_{2}) I ε_{0}$

$(ρ_{2} - ρ_{1}) I ε_{0}$

answer is D.

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Detailed Solution

By Gause law on closed surface $E_{2} S - E_{1} S = \frac{q_{B}}{ε_{0}}$

$E=ρJ = P \frac{I}{A}$

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