Force due to combination of free space and dielectric

I will make a generalized form of my question.

There are two point charges $q$, $x$ distance apart. And there is a dielectric slab of thickness $t$ and of dielectric constant $K$.

Should the force be sum of forces on them separately.

i.e. $F=\frac{k\cdot q^2}{(x-t)^2} + \frac{k\cdot q^2}{t^2*K}.$

Is this right or am I doing something wrong ?

-
 You're doing something wrong. Look at each term on the right hand side and ask yourself what it represents physically. – Alfred Centauri Jun 22 '12 at 20:55 @Alfred both represent force ? – TheGuru Jun 23 '12 at 4:20 True. What I was getting at at that each term represents the force between two particles (1) x-t apart and (2) t apart. That doesn't correspond to your physical system at all. Consider what would happen to the 2nd term as the thickness of the dielectric goes to zero. – Alfred Centauri Jun 23 '12 at 11:42 @alfred Right. Never noticed that .... Then what should be the answer ? – TheGuru Jun 23 '12 at 15:42

The correct equation, to represent the physical system, must give an answer that is bounded by the physical extremes of t = 0 and t = x, i.e.:

$\frac{kq^2}{Kx^2} < F < \frac{kq^2}{x^2}$

The lower value is when t = x and the upper value is when t = 0.

I'm thinking that we need to find an effective relative dielectric constant to use over the distance x.

$F = \dfrac{kq^2}{K_{eff}x^2}$

Given the bounds above, I'm thinking a kind of weighted harmonic mean might work.

$K_{eff} = \dfrac{x}{(x-t) + \dfrac{t}{K}}$

Note that when:

$t = 0, K_{eff} = 1$

$t = x, K_{eff} = K$

Anyhow, if you stare at this a bit, you'll see that the force, as desired, decreases as t increases and that the formula reduces to the bounds above for t = 0 and t = x.

-