Question

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?

Answer 1

Dust particles exert Force on each other F = 9.0 • 10^-2 N
The particles are moved by 1/9 th as far. We have equation for force and distance when two particles involved that is F = K q1 q2 / r^2.
So Force is inversely proportional to r^2, so the new r is r / 9, then the force becomes r^2 times the previous force.
New Force = 81 x 9.0 âś• 10^-2 = 729 x 10^-2 = 7.29 N

Answer 2

Answer: 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