Question

5. Two charges with a (charge) ratio of 1:3 are 30cm apart from each other in vacuum (free space).

   1. a) How far do both charges need to be placed in water when we want to achieve the same interaction force in both cases? (εr,water@20°C = 80.1)

Answer 1

Hey there!

We know that according to coulombs law, the force between two charges q₁ and q₂ can be given as

Given that the charges are in the ratio 1:3 with a distance of 30 cm = 0.3 m , hence

For the same force to happen underwater, it is given that the relative permittivity E_r is 80. That is

Realtive permittivity of water is E_r = E_water / E_freespace =80. Hence by substituting the same in the coulumb's law of force we get the value as 1 / 80 of the original value from that of the free space. Hence the force experienced by the charges in water would be

The condition here is F = F_water

_{Hence we get,}

Hence r = 0.011 m is the distance required to obtain the same interactive force.

I hope the explanation helps... Feel free to comment and discuss further,,, Cheers =)

1 q192

F910 0.09

9,10 80r-2 F unter =

3 0,09 8072