Electrostatic force in a medium of non uniform permittivity

[PhysicoRaj]

Homework Statement

Two point charges, $+4 μC$ and $-10 μC$ are placed 10 cm apart in air. A dielectric slab of large area and thickness 5 cm is placed between the charges. Find the force of attraction between the charges, if the dielectric has a dielectric constant of 9.

Homework Equations

$F=\frac{1}{4πε}\frac{q_1q_2}{r^2}$

$ε=ε_0{ε_r}$

The Attempt at a Solution

The electric field exists in a region of non uniform permittivity. Half of the region is free space while the other half is ε_r=9. Now should I consider the effective permittivity of the space b/n the charges? Or should I calculate the electric fields in the two different media or something like that?
Thanks.

 

[PhysicoRaj]

Well, I was searching for an equation for effective permittivity and I got it as
$ε_{eff}=\frac{ε_r}{1+ε_r\frac{L}{L_{ε_r}}}$
Plugging in 9 for $ε_r,$ 5cm each for $L$ and $L_{ε_r},$ I get $ε_{eff}=0.9$
Now
$\frac{1}{4πε}=\frac{9\times10^9}{ε_{eff}}$$=10^{10}$

So $F=\frac{10^{10}\times40\times10^{-12}}{(0.1)^2}$
$F=40N$ and its $wrong$:grumpy:

 

[PhysicoRaj]

Anybody please?

 

[Vibhor]

What is the correct answer as per answer key ?

 

[PhysicoRaj]

Ok, I have to search it, its been a long time. Wait..

 

[PhysicoRaj]

They have not given the answer.. there are 4 options: 9N, 11N, 5N, 12N.