Two charged particles exert an electrostatic force of 27 N on each other. What will the magnitude of the force be if the distance between the two charges is increased to three times the original distance?
didn't I just do this a few minutes ago?
To find the magnitude of the force when the distance between the charged particles is increased to three times the original distance, we can use Coulomb's Law. Coulomb's Law states that the electrostatic 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.
The formula for Coulomb's Law is:
F = k * (q1 * q2) / r^2
where F is the electrostatic force, k is Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.
In this scenario, the original force is given as 27 N. We'll assume that the charges of the particles remain the same.
To find the new force, we can utilize the inverse square relationship between force and distance. As the distance increases to three times the original distance, the denominator of the formula (r^2) becomes (3r)^2 = 9r^2. Therefore, the new value of r^2 becomes nine times the original value.
So, if the distance is increased to three times the original distance, the new force will be:
F' = k * (q1 * q2) / (9r^2)
Since the new force F' is what we're trying to find, we can rearrange the equation to solve for it:
F' = (k * (q1 * q2)) / (9 * r^2)
Now we can substitute the given values into the equation:
F' = (k * (q1 * q2)) / (9 * r^2)
F' = (k * (q1 * q2)) / (9 * (original r)^2)
F' = (27 N) / 9
F' = 3 N
Therefore, the magnitude of the force will be 3 N when the distance between the two charges is increased to three times the original distance.