184_notes:examples:week2_electric_field_negative_point

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184_notes:examples:week2_electric_field_negative_point [2018/05/17 16:35] – curdemma184_notes:examples:week2_electric_field_negative_point [2021/05/19 15:11] (current) – schram45
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 ===Representations=== ===Representations===
-{{ 184_notes:2_potential_negative_point.png?150 |Negative Point Charge -Q, and Point P}}+[{{ 184_notes:2_potential_negative_point.png?150 |Negative Point Charge -Q, and Point P}}] 
 + 
 +<WRAP TIP> 
 +===Assumptions=== 
 +  * Constant charge: Makes charge in electric field equation not dependent on time or space as no information is given in problem suggesting so. 
 +  * Charge is not moving: This makes our separation vector fixed in time as a moving charge would have a changing separation vector with time. 
 +</WRAP>
  
 ===Goal=== ===Goal===
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 This leaves us to find the unit vector ˆr. The first thing to do would be to draw in the separation vector, →r−Q→P. This vector points from the charge −Q to Point P since P is where we want to find the electric field (our observation location). We need to define a set of coordinate axes. We could pick the normal x- and y-axes, but this would make writing the →r and ˆr more difficult because there would be both x- and y-components to the separation vector (since it points in some diagonal direction). This leaves us to find the unit vector ˆr. The first thing to do would be to draw in the separation vector, →r−Q→P. This vector points from the charge −Q to Point P since P is where we want to find the electric field (our observation location). We need to define a set of coordinate axes. We could pick the normal x- and y-axes, but this would make writing the →r and ˆr more difficult because there would be both x- and y-components to the separation vector (since it points in some diagonal direction).
  
-{{ 184_notes:2_electric_field_negative_point_s_direction.png?300 |s-direction drawn in}}+[{{ 184_notes:2_electric_field_negative_point_s_direction.png?300 |s-direction drawn in}}]
  
 Instead, we'll pick a coordinate direction that falls along the same axis as the separation vector, →r−Q→P. Since this is a coordinate direction that we're naming, let's call this the ˆs direction. That means that →r−Q→P points in the ˆs direction, so ˆr=ˆs. Plugging this into our electric field equation gives:   Instead, we'll pick a coordinate direction that falls along the same axis as the separation vector, →r−Q→P. Since this is a coordinate direction that we're naming, let's call this the ˆs direction. That means that →r−Q→P points in the ˆs direction, so ˆr=ˆs. Plugging this into our electric field equation gives:  
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 This gives the magnitude of the electric field as |→E|=14πϵ0QR2
 This gives the magnitude of the electric field as |→E|=14πϵ0QR2
 with the direction is given by −ˆs. The electric field at P therefore points from P to the point charge. You'll find this to be true for all negative charge - **the electric field points towards negative charges**. A diagram indicating the direction of the electric field is shown below. with the direction is given by −ˆs. The electric field at P therefore points from P to the point charge. You'll find this to be true for all negative charge - **the electric field points towards negative charges**. A diagram indicating the direction of the electric field is shown below.
-{{ 184_notes:2_electric_field_negative_point_solution.png?150 |Charge Distribution Induced From Two Sides, Solution}}+[{{ 184_notes:2_electric_field_negative_point_solution.png?150 |Charge Distribution Induced From Two Sides, Solution}}]
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