The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value
Answer:
F' = (4/9)F
Explanation:
The electrostatic force between two charged objects is given by Coulomb's Law:
F = kq₁q₂/r² -------------------- equation (1)
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of second charge
r = distance between charges
Now, when the charges and distance altered as follows:
q₁' = 2q₁
q₂' = 2q₂
r' = 3r
Then,
F' = kq₁'q₂'/r'²
F' = k(2q₁)(2q₂)/(3r)²
F' = (4/9)kq₁q₂/r²
using equation (1):
<u>F' = (4/9)F</u>
