# Charge over 2 layer dielectric, image method

Physics Asked by Anthony Lethuillier on August 21, 2020

If I have a charge $$Q$$ located over a 2 layer dielectric as represented:

According to the image method: the charge $$Q’1$$ will be located at a distance $$h_1$$ under the first interface and the $$Q’2$$ will be located at a distance $$h_2$$ under the first interface.

With:

$$Q’1 = frac{epsilon_0-epsilon_1}{epsilon_0+epsilon_1} = frac{1-5}{1+5} Q$$

My problem is determining the value of $$Q’2$$, is it:
$$Q’2 = frac{epsilon_1-epsilon_2}{epsilon_1+epsilon_2} = frac{5-100}{5+100} Q$$

or

$$Q’2 = frac{epsilon_0-epsilon_2}{epsilon_0+epsilon_2}= frac{1-100}{1+100} Q$$

From Jackson's Electrodynamics: the potential in the $epsilon_1$ region for a single interface is as if there is an image charge $Q''=frac{2epsilon_1}{epsilon_1+epsilon_0}Q$ at the position of $Q$. Therefore the correct $Q'2$ is

$$Q''frac{epsilon_1-epsilon_2}{epsilon_1+epsilon_2}=frac{2epsilon_1}{epsilon_1+epsilon_0}frac{epsilon_1-epsilon_2}{epsilon_1+epsilon_2}Q$$

which seems to have the right behavior in the two limiting cases $epsilon_0=epsilon_1$ and $epsilon_1=epsilon_2$.

Answered by fferen on August 21, 2020

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