Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value. but what will happen if we decrease the distance beetween the spheres? : decreases the force of attraction or repulsion between the objects.
Answers & Comments
Answer:
Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value. but what will happen if we decrease the distance beetween the spheres? : decreases the force of attraction or repulsion between the objects.
Explanation:
:)
Verified answer
a) The electrostatic force between two charged spheres can be written as:
F = kq^2/r^2
where q is the charge on each sphere, r is the distance between their centers, and k is the Coulomb constant.
For equilibrium, the forces must balance: F_repulsive = F_attractive.
The repulsive force is given by:
F_repulsive = kq^2/r^2
The attractive force is given by:
F_attractive = kq^2/(2r)^2 = kq^2/4r^2
Setting these equal and solving for q, we get:
kq^2/r^2 = kq^2/4r^2
q^2 = 1/4q^2
q = ±1/2
So each sphere must have a charge of ±1/2.
b) Doubling the distance between spheres changes the force between them as:
F_new = kq^2/(2r)^2 = F/4
Since the forces must still balance for equilibrium, we have:
F_repulsive = F_attractive
kq^2/r^2 = F/4
Substituting the expression for F from above, we get:
kq^2/r^2 = kq^2/(4r)^2
Solving for q, we get:
q = ±1/4
So doubling the distance between the spheres reduces the required charge on each sphere by a factor of 2.