17. At what distance of separation, r, must two 5.87 x 10^-8 Coulomb charges be positioned in order for the repulsive force
between them to be 64.9 N? You must show your work for credit
To determine the distance of separation, r, between the two charges, we can use Coulomb's Law, which states that the force between two charges is given by:
F = (k * |q1 * q2|) / r^2
Where:
F is the force between the charges,
k is the electrostatic constant (9 x 10^9 N m^2 / C^2),
|q1| and |q2| are the magnitudes of the charges, and
r is the distance of separation between the charges.
In our case, both charges are 5.87 x 10^-8 Coulombs, and the repulsive force is 64.9 N. So, we have:
64.9 N = (9 x 10^9 N m^2 / C^2) * (5.87 x 10^-8 C) * (5.87 x 10^-8 C) / r^2
Now, we can rearrange the formula to solve for r:
r^2 = (9 x 10^9 N m^2 / C^2) * (5.87 x 10^-8 C) * (5.87 x 10^-8 C) / 64.9 N
r^2 = (9 x 10^9 N m^2 / C^2) * (5.87 x 10^-8 C)^2 / 64.9 N
r^2 = 3.91 x 10^-21 m^2
Taking the square root of both sides:
r = √(3.91 x 10^-21 m^2)
r ≈ 6.24 x 10^-11 meters
Therefore, the two charges must be positioned at a distance of approximately 6.24 x 10^-11 meters apart for the repulsive force between them to be 64.9 N.
64.6=8.9e9*(5.87e-8)^2/d^2
d=5.87e-8 *(sqrt (8.9e9/64.6))=
pretty close...