Question

The force between two point charges is 1.521 x 10 EXP -2 N . what will be the force when these point charges are placed in water of dielectric constant 80​

Answer 1

Therefore the force will be 1.90125 × 10¯⁴ N when these point charges are placed in water.

Given : The force between two point charges is 1.521 × 10¯² N.

To find : what will be the force when these point charges are placed in water of dielectric constant 80.

solution : We know, The electrostatic force is inversely proportional to dielectric constant of medium.

i.e., F ∝ 1/Dielectric constant

⇒F₁/F₂ = D₂/D₁

here F₁ = 1.521 × 10¯² N , D₁ = 1 [dielectric constant of vacuum/air medium is 1 ] , D₂ = 80

now, 1.521 × 10¯²/F₂ = 80/1

⇒F₂ = 1.521 × 10¯²/80

= 0.0190125 × 10¯²

= 1.90125 × 10¯⁴ N

Therefore the force will be 1.90125 × 10¯⁴ N when these point charges are placed in water.

Answer 2

Answer:

1.90125 × 10^-4N

Step-by-step explanation:

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