184_notes:linecharge

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184_notes:linecharge [2018/09/12 15:44] – [Horizontal Line of Charge] dmcpadden184_notes:linecharge [2021/07/22 18:17] (current) – schram45
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 Sections 15.1-15.2 in Matter and Interactions (4th edition) Sections 15.1-15.2 in Matter and Interactions (4th edition)
  
-[[184_notes:dist_charges|Next Page: Distributions of Charges]]+/*[[184_notes:dist_charges|Next Page: Distributions of Charges]]
  
-[[184_notes:dq|Previous Page: dQ and the →r]]+[[184_notes:dq|Previous Page: dQ and the →r]]*/
  
 ===== Lines of Charge Examples ===== ===== Lines of Charge Examples =====
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 Now we just need to fill in the pieces of this equation. Now we just need to fill in the pieces of this equation.
  
-=== Finding dQ ===+==== Finding dQ ====
  
 To find dQ, we want to split this line of charge into little chunks of charge. Since this is a linear charge distribution, we can write dQ as To find dQ, we want to split this line of charge into little chunks of charge. Since this is a linear charge distribution, we can write dQ as
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 Now we have the little bit of charge represented in terms of the little bit of length.  Now we have the little bit of charge represented in terms of the little bit of length. 
  
-=== Finding →r ===+==== Finding →r ====
  
 To find the →r, we need to write the distance from a general dQ to Point A, which in this case will only be in the ˆx direction. Remember that we can write the separation vector →r as: To find the →r, we need to write the distance from a general dQ to Point A, which in this case will only be in the ˆx direction. Remember that we can write the separation vector →r as:
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-=== Putting it together ===+==== Putting it together ====
  
 Now, we can fit all the pieces together to find the electric field by plugging in what we found for dQ, ˆr and r: Now, we can fit all the pieces together to find the electric field by plugging in what we found for dQ, ˆr and r:
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 →E=∫14πϵ0QLdx(L2+d−x)2ˆx →E=∫14πϵ0QLdx(L2+d−x)2ˆx
  
-The final piece that we need to add is limits to the integral. Since the piece of tape stretches from −L2 to L2, this means that the limits on the integral should also go from −L2 to L2 - conceptually this means that we want to add up the little bits of charge **only** along the length of the line. This gives us a final integral of:+The final piece that we need to add is limits to the integral. Since the piece of tape stretches from −L2 to L2, this means that the limits on the integral should also go from −L2 to L2 - conceptually this means that we want to add up the little bits of charge //only// along the length of the line. **This gives us a final integral of:**
 →E=∫L2−L214πϵ0QLdx(L2+d−x)2ˆx →E=∫L2−L214πϵ0QLdx(L2+d−x)2ˆx
  
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 ==== Examples ==== ==== Examples ====
-[[:184_notes:examples:Week4_charge_ring|Electric Field from a Ring of Charge]]+  * [[:184_notes:examples:Week4_charge_ring|Electric Field from a Ring of Charge]] 
 +    * Video Example: Electric Field from a Ring of Charge 
 +  * [[:184_notes:examples:Week4_charge_cylinder|Super Challenge Problem: Electric field from a Cylinder of Charge]] 
 +    * Video Example: Electric Field from a Cylinder of Charge 
 +{{youtube>I6cqhqIdG7A?large}} 
 +{{youtube>LeIp7rrwchw?large}}
  
 /*[[:184_notes:examples:Week4_charge_cylinder|Super Challenge Problem: Electric field from a Cylinder of Charge]]*/ /*[[:184_notes:examples:Week4_charge_cylinder|Super Challenge Problem: Electric field from a Cylinder of Charge]]*/
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