Question

the distance between the charged particle in air is 2 newton and when placed in medium is Er =80 then the force reduce to ?​

Answer 1

Answer:

The coulomb force between two charges q1, q2 is:

F =kq1q2/r^2

k = coulomb’s constant = 1/4πεε0

ε0 = permittivity of empty space

ε = relative permittivity of the material in which the charges are immersed

r = distance between charges

for vacuum ε=1 and for air also, to a very good approximation ε=1.

For water , at room temperature ( 20C) ε = 80.

Therefore, the force between the charges q1, q2 will be 1/80 its value in air.

Hence Fw, the force in water will be:

Fw = 80/80 = 1N.

Explanation:

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Answer 2

Answer:

force is 1 N

Explanation:

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