28.57.Model: The currents in the two segments of the wire are the same.Visualize:The electric fields rE1and rE2point in the direction of the current. Establish a cylindrical Gaussiansurface with end area athat extends into both segments of the wire.Solve:(a)Because current is conserved, I1=I2=I. The cross-section areas of the two wires are the same, so thecurrent densities are the same: J1=J2=I/A. Thus the electric fields in the two segments have strengthsEJIAEJIA11112222==σσThe electric field enters the Gaussian surface on the left (negative flux) and exits on the right. No flux passesthrough the wall of cylinder, so the net flux is Φe=E2a– E1a. The Gaussian cylinder encloses charge Qin
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 03/14/2009 for the course PHY PHYA22 taught by Professor Abbas during the Spring '09 term at University of Toronto.