How to apply Coulomb's law for a system with a dielectric slab?

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The discussion focuses on applying Coulomb's law to a system with a dielectric slab between two point charges. The initial problem involves calculating the repulsive force between the charges when a dielectric slab is introduced. Participants suggest transforming the system into an equivalent vacuum scenario by adjusting the effective separation due to the dielectric's presence. The solution involves calculating an effective separation that accounts for the dielectric's thickness and its dielectric constant. Ultimately, the user successfully solves the problem, confirming their answers match the provided key.
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Homework Statement



Two similar point charges q1 and q2 are placed at a distance r apart in the air. A dielectric slab of thickness t(<<r) having dielectric constant K is placed between the charges. Calculate the coulomb force of repulsion between the charges.

Now assume that a slab of thickness half the separation between the charges and the Coulomb's repulsive force is reduced in the ratio 9:4. Calculate K for such a slab.


Homework Equations



Coulomb's law :

F=q1q2/4πεr2

where ε=εoK



The Attempt at a Solution



I am not sure. For such a system how should I apply Coulomb's law ? We have a dielectric slab between the system. So how to make that law apply here ?

I just thought that I have to transform the given system to its equivalent vacuum one, but how should I do that ?

Please help !

Thanks in advance...:smile:
 
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Can someone help me please ? I am really clueless. :(
 
Hi Sankalp

sankalpmittal said:
I just thought that I have to transform the given system to its equivalent vacuum one, but how should I do that ?

You have the right idea .

First convert the slab width 't' into effective air separation . Can you do that ?
 
Tanya Sharma said:
Hi Sankalp



You have the right idea .

First convert the slab width 't' into effective air separation . Can you do that ?

Thanks..

If we hypothetically bring the two charges with the separation "t", then

q1q2/4πεoKt2 = q1q2/4πεox2

Solving we get x=√(K)t

Then in the actual system, effective separation is d-t+x ? Right ?

BTW, are you an IIT aspirant ?
 
sankalpmittal said:
Thanks..

If we hypothetically bring the two charges with the separation "t", then

q1q2/4πεoKt2 = q1q2/4πεox2

Solving we get x=√(K)t

Correct.

sankalpmittal said:
Then in the actual system, effective separation is d-t+x ? Right ?

r-t+x
 
Tanya Sharma said:
Correct.
r-t+x

Ya ya... I mistyped... I always do... d for distance...

Now with this information I solved both the parts completely...

Thanks a lot ! The answer matches with the key. :)
 

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