Two charged dust particles exert a force of 3.8 10^-2 N on each other. What will be the force if they are moved so they are only one third as far apart?


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Like your other question, this is an exercise in using the inverse square law.

Try it out and we'll be glad to tell you if you got it right.

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2.048

To find the new force between the charged dust particles when they are moved one third 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 charged particles is "d". According to the problem, the initial force is 3.8 \times 10^-2 N.

If the particles are moved so they are only one third as far apart, the new distance between them would be (1/3)d.

To find the new force, we can use the relationship between force and distance in Coulomb's Law:

F = k * (q1 * q2) / r^2

Where:
F is the force between the charged particles
k is Coulomb's constant (9 x 10^9 N m^2/C^2)
q1 and q2 are the charges of the particles
r is the distance between the particles

Since the charges of the particles are not given in the problem, we can assume they are equal. Let's call the charge of each particle "q".

Using these values, Coulomb's Law can be written as:

F = k * (q * q) / r^2

To find the new force, we need to substitute the new distance (1/3)d into the equation:

F_new = k * (q * q) / (1/3d)^2

Simplifying this expression, we get:

F_new = k * (q * q) / (1/9d^2)

Since the initial force is 3.8 \times 10^-2 N, we can set this value equal to F_new and solve for q:

3.8 \times 10^-2 N = k * (q * q) / (1/9d^2)

Now, we can rearrange the equation to solve for F_new:

F_new = (3.8 \times 10^-2 N) * (1/9d^2) * (9d^2 / k)

Simplifying further, we get:

F_new = (3.8 \times 10^-2 N) * (9d^2 / (9d^2 * k))

Finally, we can simplify the expression to find the new force:

F_new = (3.8 \times 10^-2 N) / k

Therefore, the new force between the charged dust particles when they are moved one third as far apart is (3.8 \times 10^-2 N)/k.