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Lindsayyyy
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Hi everyone
I have an understanding problem concerning the method of images method
Jackson uses this kind of problem in his textbook but I don't understand it.
We have a charge q on the z axis at distance d. the dielectric constant is e1 when z>=0 and e2 when z<0. Now I want to find the potential.
This is what Jackson does what I don't understand.
We use another charge q' at the opposite location to get the potential for z>0
[tex] \Phi =\frac {1}{4 \pi \epsilon_1} (\frac{q}{R_1}+\frac{q'}{R_2})[/tex]
This is equal to the case where we have only a conducting plate on the x axis. Now we have to find the potential for z<0 whereas Jackson says:
[tex] \Phi =\frac {1}{4 \pi \epsilon_2} \frac {q''}{R_1}[/tex]
I don't understand this step. The explantion which is given from Jackson doesn't help me either. How to I get to this solution?
Thanks for your help.
I have an understanding problem concerning the method of images method
Homework Statement
Jackson uses this kind of problem in his textbook but I don't understand it.
We have a charge q on the z axis at distance d. the dielectric constant is e1 when z>=0 and e2 when z<0. Now I want to find the potential.
Homework Equations
The Attempt at a Solution
This is what Jackson does what I don't understand.
We use another charge q' at the opposite location to get the potential for z>0
[tex] \Phi =\frac {1}{4 \pi \epsilon_1} (\frac{q}{R_1}+\frac{q'}{R_2})[/tex]
This is equal to the case where we have only a conducting plate on the x axis. Now we have to find the potential for z<0 whereas Jackson says:
[tex] \Phi =\frac {1}{4 \pi \epsilon_2} \frac {q''}{R_1}[/tex]
I don't understand this step. The explantion which is given from Jackson doesn't help me either. How to I get to this solution?
Thanks for your help.