Consider the junction between two uniform cylindrical wires A and B having the same diameter. Conventional current flows through the junction from A to B. As the wires must carry the same total current, the current density must have the same magnitude in each. However, because of differences in the wires’ conductivity, suppose that the electric field magnitude in A is twice that in B. What does Gauss’s law imply about the sign of the surface charge on the junction itself? (Hint: Consider the derivative of the electric field in a thin region containing the junction.)
A. The surface charge on the junction is positive.
B. The surface charge on the junction is negative.
C. The surface charge on the junction is zero.
D. Gauss’s law tells us nothing useful about the sign of the surface charge on the junction.
B. The surface charge on the junction is negative.
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