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.
Best Answer
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:
$$ F = \frac{kq_1q_2}{L^2} $$