Question

The force between two charges at a certain distance in air is unchanged when those charges are
each increased by 60% and the distance between them in a particular medium is 20% less than that
in air. Then dielectric constant of that medium is​

Answer 1

dielectric constant of medium is 4.

let q and Q are two charge seperated r distance from each other in vaccum.

from coulomb's law, F = kqQ/r² ......(1)

according to question,

each charge increased by 60% .

so, q' = q + 60% of q = 8q/5

Q = Q + 60% of Q = 8Q/5

and seperation between charges is decreased by 20% . i.e., r' = r - 20% of r = 4r/5

let dielectric constant is P. then , k' = k/P

force , F = k(8q/5)(8Q/5)/P(4r/5)²

= [4kqQ/Pr²] ........(2)

from equations (1) and (2),

kqQ/r² = 4kqQ/Pr²

⇒P = 4

hence, dielectric constant of medium is 4.

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Answer 2

$\huge\bold\purple{Answer:-}$

let q and Q are two charge seperated r distance from each other in vaccum.

from coulomb's law, F = kqQ/r² ......(1)

according to question,

each charge increased by 60% .

so, q' = q + 60% of q = 8q/5

Q = Q + 60% of Q = 8Q/5

and seperation between charges is decreased by 20% . i.e., r' = r - 20% of r = 4r/5

let dielectric constant is P. then , k' = k/P

force , F = k(8q/5)(8Q/5)/P(4r/5)²

= [4kqQ/Pr²] ........(2)

from equations (1) and (2),

kqQ/r² = 4kqQ/Pr²

⇒P = 4

hence, dielectric constant of medium is 4.