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184_notes:dq [2018/09/12 15:22] – dmcpadden184_notes:dq [2021/05/26 13:36] (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:linecharge|Next Page: Line of Charge]]+/*[[184_notes:linecharge|Next Page: Line of Charge]]
  
-[[184_notes:line_fields|Previous Page: Building Electric Field and Potential for a Line of Charge]]+[[184_notes:line_fields|Previous Page: Building Electric Field and Potential for a Line of Charge]]*/
  
 ===== dQ and the $\vec{r}$ ===== ===== dQ and the $\vec{r}$ =====
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 [{{  184_notes:dl.png?50|Vertical "little bit of length", $dy$}}] [{{  184_notes:dl.png?50|Vertical "little bit of length", $dy$}}]
  
-=== Charge on a line ===+==== Charge on a line ====
  
 For a **1D uniform charge density** (such as lines of charge), we use the variable $\lambda$, which has units of $\frac{C}{m}$ (coulombs per meter). You can calculate $\lambda$ by taking the total charge that is spread over the total length: For a **1D uniform charge density** (such as lines of charge), we use the variable $\lambda$, which has units of $\frac{C}{m}$ (coulombs per meter). You can calculate $\lambda$ by taking the total charge that is spread over the total length:
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 [{{  184_notes:da.png?200|"Little piece of area", $dA$}}] [{{  184_notes:da.png?200|"Little piece of area", $dA$}}]
  
-=== Charge on a surface ===+==== Charge on a surface ====
  
 For a **2D uniform charge density** (such as sheets of charge), we use the variable $\sigma$, which has units of $\frac{C}{m^2}$ (coulombs per meter squared). You can calculate $\sigma$ by taking the total charge that is spread over the total area: For a **2D uniform charge density** (such as sheets of charge), we use the variable $\sigma$, which has units of $\frac{C}{m^2}$ (coulombs per meter squared). You can calculate $\sigma$ by taking the total charge that is spread over the total area:
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 You can write the "little bit of area" in a variety of ways, depending on the shape of charge. You can write the "little bit of area" in a variety of ways, depending on the shape of charge.
  
-=== Charge in a volume ===+==== Charge in a volume ====
  
 Similarly, for a **3D uniform charge density** (such as a sphere of charge), we use the variable $\rho$, which has units of $\frac{C}{m^3}$ (coulombs per meter cubed). You can calculate $\rho$ by taking the total charge that is spread over the total volume: Similarly, for a **3D uniform charge density** (such as a sphere of charge), we use the variable $\rho$, which has units of $\frac{C}{m^3}$ (coulombs per meter cubed). You can calculate $\rho$ by taking the total charge that is spread over the total volume:
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 Because we talk about lines of charge, we usually pick some length variable like "L", "x" or "y". **You always want this variable to match the "little bit of length" variable that you chose for your dQ.** So if you choose a "dL", you want to use "L" as your variable; if you choose "dx", you want to use "x"; etc.  This will prevent you from referring to the same length with two different variable names. Because we talk about lines of charge, we usually pick some length variable like "L", "x" or "y". **You always want this variable to match the "little bit of length" variable that you chose for your dQ.** So if you choose a "dL", you want to use "L" as your variable; if you choose "dx", you want to use "x"; etc.  This will prevent you from referring to the same length with two different variable names.
  
-For the picture shown, we can find the $\vec{r}$ by splitting it into components. First, we need to pick a coordinate system - so lets pick the $(0,0)$ location to be at the bottom of the tape with +x being to the right and +y being up like normal. The x component here will always have a magnitude of length d; no matter which r vector we pick, the $|r_x|=d$. However, it points in the -x direction, so $r_x=-d \hat{x}$. The y-component here is a little more tricky. If we start by saying we want the magnitude of $r_y$, then we can focus first on the length then deal with the direction later. Remember we want to find the y component of r, which points from the dQ to the point of interest. So we define our variable, lets say "y" because the tape is hanging vertically, to be the length between zero and where ever the dQ is. Then $r_y$ is the length of tape that is //not// covered by y, so $|r_y|=L-y$ where L is the total length of the tape. This component of the vector should always point up, so then we get $\vec{r}_y=(L-y) \hat{y}$. When we combine these components, we get the total r vector:+For the picture shown, we can find the $\vec{r}$ by using the same separation vector equation that we were using before: 
 +$$ \vec{r} = \vec{r}_{observation}-\vec{r}_{source}$$ 
 +First, we need to pick a coordinate system - so lets pick the $(0,0)$ location to be at the bottom of the tape with +x being to the right and +y being up like normal. In this coordinate system, the observation location is at a height of L and a distance of d in the -x direction. So we get: 
 +$$\vec{r}_{obs} = \langle -d, L, 0 \rangle$$ 
 +The source location is a little more tricky because our dQ could be any on the piece of tape. For example, dQ could be at the bottom of the piece of tape, at the top, or somewhere in the middle - so we ideally we want to write the location of dQ (aka the $\vec{r}_{source}$) so that it is in terms of some variable that we can adjust. Since the tape is oriented vertically, we'll pick a variable $y$ for the height of the dQ. This means we can write $\vec{r}_{source}$ as: 
 +$$\vec{r}_{source}= \langle 0, y, 0 \rangle$$ 
 +The x-component of $\vec{r}_{source}$ here is zero because we have the tape located on the y-axis, which would be true no matter where on the tape our dQ is located. When we combine these pieces, we get the total separation vector: 
 +$$\vec{r}=\langle -d, L, 0 \rangle - \langle 0, y, 0 \rangle$$
 $$\vec{r}=\langle -d, L-y, 0 \rangle$$ $$\vec{r}=\langle -d, L-y, 0 \rangle$$
-This way of writing the $\vec{r}$ works for any spot along the piece of tape, and functions like any other vector (we can find its magnitude, unit vector, etc.). Because $\vec{E}$ and $V$ rely heavily on $\vec{r}$ and the $|r|$, we will use this method and reasoning when we are dealing with lines of charge (though this works more generally for planes, spheres, or blobs too).+This equation for the $\vec{r}$ works for any spot along the piece of tape, and functions like any other vector (we can find its magnitude, unit vector, etc.). Because $\vec{E}$ and $V$ rely heavily on $\vec{r}$ and the $|r|$, we will use this method and reasoning when we are dealing with lines of charge (though this works more generally for planes, spheres, or blobs too).
  
 ====Examples==== ====Examples====
-[[:184_notes:examples:Week4_tilted_segment|A Tilted Segment of Charge]]+  * [[:184_notes:examples:Week4_tilted_segment|A Tilted Segment of Charge]] 
 +  * [[:184_notes:examples:Week4_two_segments|Two Segments of Charge]] 
 +    *  Video Example: Two Segments of Charge 
 +{{youtube>BiqTwMrD774?large}}
  
-[[:184_notes:examples:Week4_two_segments|Two Segments of Charge]] 
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