Question

Two point charges placed in a medium of dielectric constant 5 are at a

distance r between them, experience an electrostatic force ‘F’. The

electrostatic force between them in vacuum at the same distance r will

be-

(i) 5F

(ii) F

(iii) F/2

(iv) F/5​

Answer 1

Given info : Two point charges placed in a medium of dielectric constant 5 are at a  distance r between them, experience an electrostatic force F.

To find : the electrostatic force between them in vacuum at the same distance r will be...

solution : let two charge Q₁ and Q₂ are placed in a medium of dielectric constant, K = 5, distance between them is r.

from Coulomb's law, force between charges will be , F = $\frac{1}{4\pi\epsilon_0K}\frac{Q_1Q_2}{r^2}$ ...(1)

now the same charges are placed in vacuum and also the seperation between them is same.

force between them will be , F' = $\frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{r^2}$ ...(2)

diving equation (2) by equation (1) we get,

⇒ F'/F = K/1

⇒ F' = KF = 5F [∵ K = 5 ]

therefore the electrostatic force between them in vacuum will be 5F.

Answer 2

Answer:

i)5F

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