EOC_Solution_28_57 - 28.57. Model: The currents r the two...

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28.57. Model: The currents in the two segments of the wire are the same. Visualize: The electric fields r E 1 and r E 2 point in the direction of the current. Establish a cylindrical Gaussian surface with end area a that extends into both segments of the wire. Solve: (a) Because current is conserved, I 1 = I 2 = I . The cross-section areas of the two wires are the same, so the current densities are the same: J 1 = J 2 = I / A . Thus the electric fields in the two segments have strengths E J I A E J I A 1 1 11 2 2 22 == σσ The electric field enters the Gaussian surface on the left (negative flux) and exits on the right. No flux passes through the wall of cylinder, so the net flux is Φ e = E 2 a E 1 a . The Gaussian cylinder encloses charge Q in
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