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

 

[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.