Two charged dust particles exert a force of 5.3 x 10^-2 N on each other. What will be the force if they are moved so they are only one-ninth as far apart?
12.5
7.2e-2
To find the new force between the charged dust particles when they are moved one-ninth as far apart, we can use Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's assume the initial distance between the particles is represented by "d", and the initial force between them is represented by "F". We are given that the initial force F is 5.3 x 10^-2 N.
Now, let's represent the new distance between the particles as "d/9" since they are moved one-ninth as far apart. We want to find the new force, which we'll represent as "F'".
Using Coulomb's Law, we can write the proportionality equation:
F ∝ (q1 * q2) / d^2
where q1 and q2 are the charges of the particles.
Since we are comparing the new force (F') to the initial force (F) and keeping the charges the same, we can write:
F' / F = [(q1 * q2) / (d/9)^2] / [(q1 * q2) / d^2]
Since the charges cancel out, we can simplify further:
F' / F = (d^2 / (d/9)^2)
F' / F = (d^2) / [(d^2) / 81] = 81
Now, we can solve for F':
F' = 81 * F
Plugging in the given initial force (F = 5.3 x 10^-2 N):
F' = 81 * (5.3 x 10^-2 N)
Calculating this expression, we find:
F' = 4.293 N
Therefore, the new force between the charged dust particles will be 4.293 N when they are moved one-ninth as far apart.