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184_notes:examples:week4_charge_cylinder [2018/02/05 17:33] – tallpaul | 184_notes:examples:week4_charge_cylinder [2021/06/29 18:10] – schram45 |
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</WRAP> | </WRAP> |
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Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ | |
Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ | |
Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ | |
Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ | |
Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ | |
Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ | |
Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ | |
Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ | Notice that our $\text{d}Q$ is different than other $\text{d}Q$s we have used so far. But using a whole ring as our $\text{d}Q$ makes sense. The cylindrical shell is 2-dimensional, which means that if our $\text{d}Q$ 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 $\text{d}Q$ 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): $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$$ |
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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*} |
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| Since $z$ is very large we will once again eliminate any constant terms tied in with it.$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}\hat{x}$$ |
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| 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. |