184_notes:examples:week4_charge_cylinder

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184_notes:examples:week4_charge_cylinder [2018/02/03 21:50] – tallpaul184_notes:examples:week4_charge_cylinder [2021/06/29 18:10] – schram45
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   * Slice the cylindrical shell into thin rings, which we know about from the [[184_notes:examples:week4_charge_ring|previous example]].   * Slice the cylindrical shell into thin rings, which we know about from the [[184_notes:examples:week4_charge_ring|previous example]].
   * Write an expression for $\text{d}Q$, which is the charge of one of the rings.   * Write an expression for $\text{d}Q$, which is the charge of one of the rings.
-  * Decided o+  * Decide on a consistent way to define the location of the ring, and use this to write an expression for $\text{d}Q$
   * Assign a variable location to the $\text{d}Q$ piece, and then use that location to find the separation vector, $\vec{r}$.   * Assign a variable location to the $\text{d}Q$ piece, and then use that location to find the separation vector, $\vec{r}$.
   * Write an expression for $\text{d}\vec{E}$.   * Write an expression for $\text{d}\vec{E}$.
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 Now that we are okay on defining $\text{d}Q$ and $\vec{r}$, we update the representation to reflect these decisions: Now that we are okay on defining $\text{d}Q$ and $\vec{r}$, we update the representation to reflect these decisions:
 {{ 184_notes:4_cylinder_dq.png?400 |Cylindrical Shell dQ}} {{ 184_notes:4_cylinder_dq.png?400 |Cylindrical Shell dQ}}
 +text etgdsygt fzuhf gfsfzubfuzsuf z xd uyfg cgi.
 +kki99ki.
 +
  
 Trivially, we also have $r=|\vec{r}|=|z-x|$. We retain the absolute value notation, so that we can generalize for when $P$ is inside the cylinder. You see below that the absolute value notation immediately drops out. Trivially, we also have $r=|\vec{r}|=|z-x|$. We retain the absolute value notation, so that we can generalize for when $P$ is inside the cylinder. You see below that the absolute value notation immediately drops out.
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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\epsilon_0Lz}\left(\frac{\frac{L}{z}}{1-\frac{L^2}{4z^2}}\right) \\+        &= \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.$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}\hat{x}$$ 
 + 
 +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.
  
  • 184_notes/examples/week4_charge_cylinder.txt
  • Last modified: 2021/07/22 18:21
  • by schram45