184_notes:linecharge

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revisionBoth sides next revision
184_notes:linecharge [2021/02/13 19:13] – [Putting it together] bartonmo184_notes:linecharge [2021/02/13 19:13] – [Putting it together] bartonmo
Line 46: Line 46:
 $$\vec{E}=\int\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$ $$\vec{E}=\int\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$
  
-The final piece that we need to add is limits to the integral. Since the piece of tape stretches from $-\frac{L}{2}$ to $\frac{L}{2}$, this means that the limits on the integral should also go from $-\frac{L}{2}$ to $\frac{L}{2}$ - conceptually this means that we want to add up the little bits of charge //only// along the length of the line. This gives us a final integral of:+The final piece that we need to add is limits to the integral. Since the piece of tape stretches from $-\frac{L}{2}$ to $\frac{L}{2}$, this means that the limits on the integral should also go from $-\frac{L}{2}$ to $\frac{L}{2}$ - conceptually this means that we want to add up the little bits of charge //only// along the length of the line. **This gives us a final integral of:**
 $$\vec{E}=\int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$ $$\vec{E}=\int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$
  
  • 184_notes/linecharge.txt
  • Last modified: 2021/07/22 18:17
  • by schram45