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If two point charges $q_{1}$ and $q_{2}$ are seperated by a distance $l$ apart, and the space between them is filled with a variable dielectric constant. Near $q_{1}$, the dielectric constant is $K_{1}$, and near $q_{2}$, it is $K_{2}$ while increasing linearly with distance. Then I need to find the interaction force between the two charges.

I saw this answer, but I don't understand how I can use that result for my problem. I assume the answer involves integration(which I know), since the dielectric constant is varying.

Please, I don't need a solution, just a brief explanation of how to go about this.

Community
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Sharat
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2 Answers2

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Take the distance between the charges as $D$, and we'll assume that the dielectric constant, $K$, at a distance $x$ between the charges is given by some function $f(x)$. Actually let's assume that $f(x)$ gives $\sqrt{K}$ as this makes the notation clearer.

If you take some small element $dx$, then the effective length of this element is $\sqrt{K}\, dx = f(x)\, dx$. We get the total effective length by integrating to add up all the $dx$s, so the total effective length $L$ is:

$$ L = \int_0^D f(x)\, dx $$

And that's it. You need to work out what function $f(x)$ gives you $\sqrt{K}$ at a distance $x$ between the charges, then do the integration to work out $L$. Finally, the force is just:

$$ F = \frac{kq_1q_2}{L^2} $$

John Rennie
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If the charges are seperated by dielectric slabs of fixed dielectric constants that is easy to solve by calculating the effective length of dielectric in vacuum.But if it some function of distance then we need to know the Function first to find the effective distance in vacuum.

bhatt physics
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