184_notes:examples:week6_charges_circuit

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184_notes:examples:week6_charges_circuit [2017/09/26 14:25] – [Example: Charge Distribution on the Bends of a Circuit] tallpaul184_notes:examples:week6_charges_circuit [2021/06/08 00:41] (current) schram45
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 +[[184_notes:q_in_wires|Return to Surface Charges around a Circuit]]
 +
 =====Example: Charge Distribution on the Bends of a Circuit===== =====Example: Charge Distribution on the Bends of a Circuit=====
 On the circuit shown below, draw how you would expect charge to distribute on the surface of the wire near the bends in the circuit. On the circuit shown below, draw how you would expect charge to distribute on the surface of the wire near the bends in the circuit.
  
-{{ 184_notes:6_bends.png?300 |Circuit with Bends}}+[{{ 184_notes:6_bends.png?300 |Circuit with Bends}}]
  
 ===Facts=== ===Facts===
-  * Electric field is constant in the wire.+  * Electric field is constant in the wire - created by the surface charges on the wire, not the battery.
   * Electric field points along the direction of the wire, from positive to negative.   * Electric field points along the direction of the wire, from positive to negative.
   * Charges can build up on the surface of the wire.   * Charges can build up on the surface of the wire.
  
-===Lacking=== +===Goal=== 
-  * Charge distribution near the bends in the wire+  * Draw charge distribution near the bends in the wire.
- +
-===Approximations & Assumptions=== +
-  * There are no external electric fields.+
  
 ===Representations=== ===Representations===
   * We represent the circuit as in the example statement above.   * We represent the circuit as in the example statement above.
   * We will represent the charge distribution with little pluses ($+$) and minuses ($-$).   * We will represent the charge distribution with little pluses ($+$) and minuses ($-$).
 +
 ====Solution==== ====Solution====
-Firstnotice that we probably do not want to do any calculations heresince the it will not be fun to take dot-product of the dipole's electric field and the area-vectorand it will get very messy very quickly when we start integrating over the surface of the cylinderInstead, we evaluate the situation more qualitativelyConsider the electric field vectors of the dipole near the surface of the cylinder: +<WRAP TIP> 
-{{ 184_notes:5_dipole_field_lines.png?300 |Dipole Electric Field Lines}}+=== Assumption === 
 +In order to talk about individual charged particles in the circuit, we will have to make an assumption about the charged particles which are actually moving, the ones we think about when we think about "current"
 +  * The mobile charge carriers are positive. 
 +It doesn't so much matter what we choose, but we want to be clear about the assumption, so that our discussion take place with this in mind, and so our reasoning will be consistent. 
 + 
 +We will also assume that we have a perfect battery in this case to drive steady current and provide excess charges to be distributed on the wire. 
 +</WRAP> 
 + 
 +Note that we are only looking for the surface charges on the bends here. We need to figure out a way to distribute them so that the electric field stays constant in magnitude and along the direction of the wire. Let'consider the nature of the electric field near the positive end of the battery. In the wire, it will be pointing left into the bend. It will also have a large magnitude since it is so close to the batteryTo reduce the magnitude, and ensure that the electric field points down after the bend, we place a lot of positive surface charges on the outer edge of the bend. 
 + 
 +What would have happened if we did not place these charges here? Conventional current would follow the electric field. An electric field vector near the bend would point left, and instead of following the wire, positive charge would build up along the outer surface of the bend (the wider part of the bend) in the wire. Charge would continue to build up until we reach a point where an electric field vector near the bend points down! See below for a representation of how the surface charges near the bends in the wire would look. The electric field (and direction of conventional current) is shown with arrows.
  
-Notice that the vectors near the positive charge are leaving the cylinder, and the vectors near the negative charge are enteringNot only this, but they are mirror images of each other. Wherever an electric field vector points out of the cylinder on the right side, there is another electric field vector on the left that is pointing into the cylinder at the same angle. These mirror image vectors also have the same magnitude, though it is a little tougher to visualize.+[{{ 184_notes:6_bends_charges.png?300 |Circuit with Surface Charges on the Bends}}]
  
-We could write this as a comparison between the left and right side of the cylinderThe $\text{d}\vec{A}$-vectors are mirrored for left vs. rightand the $\vec{E}$-vectors are mirroredbut opposite $\left(\vec{E}_{left}=-\vec{E}_{right}\right)$We tentatively write the equality, +Notice that as you traverse the length of the wire, the surface charges become less positive and more negative, and eventually start to accumulate on the tighter part of the bendWe require each bend to be more negative than the previous oneso that the electric field on the straight parts of the wire points in the desired direction. And we start to draw the bends with negative charge on the tighter partbecause all we need is for the wider part of the bend to be more positive than the tighter partWhen drawing positive charge, it will be on the wider part. When drawing negative chargeit will be on the tighter part.
-$$\Phi_{left}=-\Phi_{right}$$+
  
-Putting it together, we tentatively write: +<WRAP TIP> 
-$$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$ +=== Note === 
-We gain more confidence when we read the [[184_notes:gauss_ex|next section of notes]], where we define "Gauss' Law". This law states that the total flux through a close surface is the amount of charge enclosed divided by $\epsilon_0$, the [[https://en.wikipedia.org/wiki/Vacuum_permittivity|permittivity of free space]]. +The accumulation of charge around the bends might look slightly different if we had chosen for the mobile charge carriers to be //negative// in our assumption at the top of the solution. It's a worthwhile exercise to go through the reasoning again with the flipped assumption. 
-$$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ +</WRAP>
-Since the total charge of the dipole is $0$, then indeed the charge enclosed is $0$, and we were correct with our reasoning about the electric field and flux above.+
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