Question

When two-point charges are placed some distance apart in vacuum experience a force of 800 N, what will be the force between charges in a medium of dielectric constant 80? *​

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.