Timeline for Method of images involving a charged wire and two different dielectric materials filling all of space
Current License: CC BY-SA 4.0
5 events
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| Aug 20, 2020 at 15:50 | comment | added | Laff70 | I hope this site[ photonics101.com/dielectrics/… ] provides a good explanation. For an observer in x>0, $k=\frac{1}{4\pi\epsilon_1}$. For and observer in x<0, $k=\frac{1}{4\pi\epsilon_2}$. Use the same k for all observed charge. | |
| Aug 20, 2020 at 15:01 | comment | added | Ofek Aman | Hmmm, they indeed got a somewhat different result... Can you explain how you obtained these charge densities? Also, after finding the charge density, what dielectric coefficient will appear in the expression for the fields originating from the respective wires? Will it be the respective dielectric constants of the regions where you put the image wires? | |
| Aug 20, 2020 at 14:23 | comment | added | Laff70 | I don't know why they'd ever use $\epsilon_0$ with how you described the problem. The image charge seen from x>0 should have charge density $\lambda\cdot\frac{\epsilon_1-\epsilon_2}{\epsilon_1+\epsilon_2}$ whereas the other image charge should have charge density $\lambda\cdot\frac{2\epsilon_2}{\epsilon_1+\epsilon_2}$. There are three possible explanations, $\epsilon_2=\epsilon_0$, $\epsilon_2=\epsilon_0\epsilon_r$, or they made a mistake. | |
| Aug 20, 2020 at 7:29 | comment | added | Ofek Aman | My course just published the answers to this and I've seen that their answer is just as you say. However, I don't understand why would the field take such a form? Also, I saw that in the expression for the field they left the field of the original wire with $\epsilon_1$ instead of $\epsilon_0$, whereas in the expression for the field of the image wires there was $\epsilon_0$ (in both cases, x>0 and x<0). What is the reason for that? | |
| Aug 20, 2020 at 1:21 | history | answered | Laff70 | CC BY-SA 4.0 |