Two charged dust particles exert a force of 9.0 ✕ 10-2 n on each other. what will be the force if they are moved so they are only one-ninth as far apart?

The new force would be F' = 7.29x10 N

Explanation: We know that the force between the charges is:

9.0x10^-2 N

Now, the force between two charged particles can be obtained by the Coulomb's law:

F = k*q1*q2/r^2

where k is a constant, q1 and q2 are the charges of the particles and r is the distance between the particles.

then we have that:

k*q1*q2/r^2 = 9.0x10^-2 N

if we move tha particles only one ninth as far apart, then we have a new force:

F´= k*q1*q2/(r/9)^2 = 181k*q1*q2/r^2 = 81*F = 81*9.0x10^-2 N = 729x10^-2

F' = 7.29x10 N

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .