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--- 184_notes:examples:week7_wire_dimensions [2017/10/04 13:41] – [Solution] tallpaul
+++ 184_notes:examples:week7_wire_dimensions [2017/10/04 13:53] – [Solution] tallpaul
@@ Line 24: / Line 24: @@
 ====Solution====
-Let's start with segment 1. The power dissipated through the segment is just $$P_1=I_1 \Delta V_1 = 5 \text{ W}$$
+Let's start with segment 1. The electric field is constant since the wire is uniform with respect to the rest of the segment, so we get $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ V/m}$$
-The electric field is constant since the wire is uniform with respect to the rest of the segment, so we get $$E_1 = \frac{\Delta V_1}{L_1} = 0.5 \text{ V/m}$$
+The power dissipated through the segment is just $$P_1=I_1 \Delta V_1 = 5 \text{ W}$$
+Now, for segment 2. We can use [[184_notes:r_energy#Conservation_of_Charge_in_Circuits|what we know]] about charge in steady state circuits to determine the electric field: $$E_2=\frac{A_1}{A_2}E_1 = \frac{\pi (d_1/2)^2}{\pi (d_2/2)^2}E_1=9E_1=4.5\text{ V/m}$$