184_notes:examples:week4_charge_cylinder

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184_notes:examples:week4_charge_cylinder [2018/02/05 17:33] – tallpaul184_notes:examples:week4_charge_cylinder [2021/07/22 18:21] (current) – schram45
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 ===Goal=== ===Goal===
   * Find the electric field at P.   * Find the electric field at P.
 +
 +===Assumptions===
 +  * This is a perfect cylinder: This simplifies down the geometry of the problem and allows us to use any equations related to the geometry of the cylinder
 +  * There is no top or bottom: We make this assumption so that we can break the cylinder up into rings and not have to do anything with the top or bottom sides of the cylinder (essentially we are dealing with a 2D tube).
  
 ===Representations=== ===Representations===
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 </WRAP> </WRAP>
  
-Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL 
-Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL 
-Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL 
-Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL 
-Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL 
-Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL 
-Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL 
 Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL Notice that our dQ is different than other dQs we have used so far. But using a whole ring as our dQ makes sense. The cylindrical shell is 2-dimensional, which means that if our dQ is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our dQ as a fraction of the total Q based on the thickness of our ring (we set our coordinates such that +x-direction is to the right, and the center of the cylinder is at the origin): dQ=λdl=QdxL
  
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 \vec{E} &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{1}{1-\frac{L}{2z}}-\frac{1}{1+\frac{L}{2z}}\right) \\ \vec{E} &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{1}{1-\frac{L}{2z}}-\frac{1}{1+\frac{L}{2z}}\right) \\
         &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{\left(1+\frac{L}{2z}\right)-\left(1-\frac{L}{2z}\right)}{\left(1-\frac{L}{2z}\right)\left(1+\frac{L}{2z}\right)}\right) \\         &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{\left(1+\frac{L}{2z}\right)-\left(1-\frac{L}{2z}\right)}{\left(1-\frac{L}{2z}\right)\left(1+\frac{L}{2z}\right)}\right) \\
-        &= \frac{Q\hat{x}}{4\pi%5}+        &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{\frac{L}{z}}{1-\frac{L^2}{4z^2}}\right) = \frac{Q\hat{x}}{4\pi\epsilon_0z^2}\left(\frac{1}{1-\frac{L^2}{4z^2}}\right) \\ 
 +        &= \frac{Q\hat{x}}{4\pi\epsilon_0\left(z^2-\frac{L^2}{4}\right)} 
 +\end{align*} 
 + 
 +Since z is very large we will once again eliminate any constant terms tied in with it.→E=14πϵ0Qz2ˆx 
 + 
 +As we can see this is exactly the equation we get for a point charge! This should be expected. When viewing charged objects from far away they can be approximated as points, kinda like looking at a person from a distance. 
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  • Last modified: 2018/02/05 17:33
  • by tallpaul