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Force btwn 2 identical charges placed at a distance r in vacuum is F . Now a slab of dielectric constant 4is inserted btwn these 2 charges . If the thickness of the slab is r/2 then the force btwn the charges will become ,?
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Force btwn 2 identical charges placed at a distance r in vacuum is F ....
Introduction:
The force between two identical charges in vacuum is given by Coulomb's law, F = k(q^2/r^2), where F is the force, k is the electrostatic constant, q is the charge, and r is the distance between the charges. When a dielectric slab is inserted between the charges, the force between them is modified due to the presence of the dielectric material.

Explanation:
Step 1: Electric Field without Dielectric
When the charges are placed in vacuum, they create an electric field between them. The electric field is given by E = k(q/r^2). The force between the charges is then given by F = qE.

Step 2: Electric Field with Dielectric
When a dielectric slab is inserted between the charges, it affects the electric field. The presence of the dielectric material polarizes the charges within it, creating an additional electric field that opposes the original electric field.

The electric field within the dielectric is reduced by a factor of the dielectric constant (ε) of the material. So, the electric field within the dielectric becomes E' = E/ε.

Step 3: Force with Dielectric
The force between the charges in the presence of the dielectric can be calculated using the modified electric field. The force is given by F' = qE'.

Step 4: Applying the Given Parameters
In this case, the dielectric constant of the slab is given as 4 and the thickness of the slab is r/2. As the slab is inserted between the charges, the distance between them is effectively reduced by the thickness of the slab, i.e., r' = r - r/2 = r/2.

Step 5: Calculating the Force
Using the formula for the electric field without the dielectric, we have E = k(q/r^2). Substituting the given parameters, the electric field becomes E = k(q/(r/2)^2) = 4k(q/r^2).

Using the formula for the force with the dielectric, we have F' = qE'. Substituting the calculated electric field, the force becomes F' = q(4k(q/r^2)/4) = k(q^2/r^2).

Conclusion:
Therefore, the force between the charges when a slab of dielectric constant 4 is inserted between them, with a thickness of r/2, remains the same as the force in vacuum. The presence of the dielectric material alters the electric field but does not affect the force between the charges.
Community Answer
Force btwn 2 identical charges placed at a distance r in vacuum is F ....
Let the charge be q.
Given, distance = r/2
Force = F
Dielectric constant , k = 4
F = kq^2/r^2
F = 4q^2/(r^2/4)
F = 16q^2/r^2.
When the thickness becomes half and dielectric constant 4 is inserted, force exerted will be 16 times the first given force F.
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Force btwn 2 identical charges placed at a distance r in vacuum is F . Now a slab of dielectric constant 4is inserted btwn these 2 charges . If the thickness of the slab is r/2 then the force btwn the charges will become ,?
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