Two charged bodies exert a force of 0.390 N on each other. If they are moved so that they are one-third as far apart, what force is exerted?
is it .390 times 3??
F1 = constant/d^2
F2 = same constant/ (d/3)^2 = constant *9/d^2
F2/F1 = 9
so nine times the original force if 1/3 of original seperation
To determine the force exerted when the two charged bodies are one-third as far apart, we need to apply the inverse square law of electric force.
The electric force between two charged bodies is given by the equation:
F = k * (q1 * q2) / r^2
Where:
- F is the force between the two charged bodies
- k is the electrostatic constant (k ≈ 9 × 10^9 N*m^2/C^2)
- q1 and q2 are the magnitudes of the charges of the two bodies
- r is the distance between the two bodies
In this case, we are given that the force between the two bodies is 0.390 N. Let's assume this is the original force (F1) when the bodies are at a certain distance apart (r1).
F1 = 0.390 N
Now, if they are moved so that they are one-third as far apart, the new distance (r2) will be one-third of the original distance (r1).
r2 = (1/3) * r1
To find the new force (F2), we can use the fact that the force is inversely proportional to the square of the distance (F ∝ 1/r^2):
F1 / F2 = (r2 / r1)^2
Substituting the values we know:
0.390 N / F2 = [(1/3) * r1 / r1]^2
Simplifying:
0.390 N / F2 = (1/3)^2
0.390 N / F2 = 1/9
F2 = 9 * 0.390 N
F2 ≈ 3.51 N
Therefore, when the charged bodies are one-third as far apart, the force exerted between them is approximately 3.51 N.
So, the answer is not 0.390 times 3. The new force is approximately 3.51 N, which is 9 times greater than the original force.