Answer:
They are now 3.75 cm apart.
Explanation:
It is given that, two small charged spheres are 6.52 cm apart. It they are moved, the force is found to be become tripled.
The electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Let initial force is F' so,
[tex]F'=k\dfrac{q_1q_2}{(0.0652)^2}[/tex]...........(1)
When they moved apart, let final force is F'' = 3 F'.
[tex]3F'=k\dfrac{q_1q_2}{r'^2}[/tex]...................(2)
Taking ratios of equation (1) and (2) :
[tex]r'^2=\dfrac{(0.0652)^2}{3}[/tex]
[tex]r'=0.0014\ m[/tex]
or [tex]r'=0.0375\ m[/tex]
r' = 3.75 cm
Hence, this is the required solution.
This question can be solved using the formula of electrostatic forces from Colomb's Law.
The charged spheres are now "3.8 cm" apart.
The formula for the electrostatic force is given by Colomb's Law as follows:
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where,
F = Electrostatic Force
k = Colomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of second charge
r = distance between the charges = 6.52 cm = 0.0652 m
Therefore, using values:
[tex]F = \frac{kq_1q_2}{(0.0652\ m)^2} ---------------- eqn(1)[/tex]
Now, the force becomes 3 times of the initial value, with the same magnitude of charges:
[tex]F' = 3F\\\\using\ eqn\ (1):\\\\\frac{kq_1q_2}{r'^2}=3\frac{kq_1q_2}{(0.0652\ m)^2}\\\\r'^2=\frac{(0.0652\ m)^2}{3}\\\\r' = \sqrt{0.00141\ m^2}[/tex]
r' = 0.038 m = 3.8 cm
The attached picture illustrates Colomb's Law.
Learn more about Colomb's Law here:
https://brainly.com/question/506926?referrer=searchResults
