The electrostatic force of attraction between two small spheres that are 1.0 meter apart is F. If the distance between the spheres is decreased to 0.5 meter, the electrostatic force will be
a) F/2
b) 2f
c) F/4
d) 4F
I know when the distance is decreased the Fe would decrease as well.
To calculate the electrostatic force between two small spheres, we can use Coulomb's Law. According to Coulomb's Law, the electrostatic force (Fe) between two point charges is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them:
Fe = k * (q1 * q2) / r^2
Where k is the electrostatic constant. However, in this question, we are considering two small spheres with equal charges, so q1 and q2 can be considered the same. Therefore, we can simplify Coulomb's Law equation to:
Fe = k * (q^2) / r^2
Now, let's consider the ratio between the initial distance (r1 = 1.0 meter) and the final distance (r2 = 0.5 meter). The ratio between r1 and r2 is:
r1/r2 = 1.0/0.5 = 2
To determine the effect of this change in distance on the electrostatic force, we need to analyze how the force changes with respect to the distance. Since the force is inversely proportional to the square of the distance, when the distance is halved (decreased by a factor of 2), the force will be increased by a factor of (2^2) = 4.
Therefore, the correct answer is (d) 4F.