Question

Two point charges are placed at separation d in vacuum and the force between them is F. Now a dielectric slab of thickness t = d/3 and dielectric constant K is placed
between the charges and the force becomes 9F/25. Find the
value of K.​

Answer 1

Answer:

`F prop (1)/(d^(2))` …(i)

`F' prop (1)/(((2d)/(3) + (d)/(3) sqrt(K))^(2))` ….(ii)

`(F')/(F) = (9)/(25) = (1)/(((2)/(3) + (sqrt(K))/(3))^(2))`

`(2)/(3) + (sqrt(K))/(3) = (5)/(3)`

`K = 9`

Explanation : Let two points charges `q_(1), q_(2)` be placed in a medium of dielectric constant `K` at separation `r`.

`F_(1) = (1)/(4 pi in_(0)) (q_(1) q_(2))/(K r^(2))` (i)

If we want that charges should be placed in vacuum to have the same force as in medium, we have to increase separation between charges.

`F_(1) = (1)/(4 pi in_(0)) (q_(1) q_(2))/(r'^(2))` .....(ii)

Since `F_(1) = F_(2)`

`r' = r sqrt(K)`

When the slab of thickness `t` is placed between charges , `t` is equivalent to `t sqrt(K)` of vaccum.

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