two point charges placed at a distance of 20cm in air repel each other with a certain force.when a dielectric slab of thickness 8cm and dielectric constant K is introduced between these point charges,force of interaction becomes half of its previous value.what is the approximate value of K?
a.2
b.under root 2
c.4
d.1

Question

two point charges placed at a distance of 20cm in air repel each other with a certain force.when a dielectric slab of thickness 8cm and dielectric constant K is introduced between these point charges,force of interaction becomes half of its previous value.what is the approximate value of K?
a.2
b.under root 2
c.4
d.1

anonymous · Answer

F=(1/4pi *K* epislon)*(q1q2/r2) if K=2 then F would be half

naveenbabbar · Answer

Here you can use Coulomb's law but firstly you have to find equivalent distance in space.
Use the formula
$$F=k q _{1}q _{2}divr ^{2}and F/2=k q _{1}q _{2}\div[length of \space \between charges without dielectric+thickness of the slab(underroot of dielectric constant-1)]Squared$$
This gives you answer approximately 4 i,e; c option.

anonymous · Answer

@naveenbabbar  i didnt get it

anonymous · Answer

@fazeelayaz how?

anonymous · Answer

hello guizz my name is ahmed yaseen gaad its proper solution is=
first its formula is f=q1 q2/4piepslon not (r-t +t under root k) now compare f1 with f2 in this u get that q1 q2 /4piepslonnot is canceld and remained only r-t+tunderroot k from question f2 = f/2 
then final formula will be 2=r-t+tunderrootk 
note t for thickness of dilectric slab thanku 
guizz this is my easiest question this is my email address in wich u can send me toughest questions i will answer u every one of them = samigaad.yg@gmail.com