Answer:
[tex]3.2\cdot 10^{-8} N[/tex]
Explanation:
The inital electrostatic force between the two spheres is given by:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
[tex]F=8\cdot 10^{-9} N[/tex] is the initial force
k is the Coulomb's constant
q1 and q2 are the charges on the two spheres
r is the distance between the two spheres
The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have
[tex]r'=\frac{r}{2}[/tex]
Therefore, the new electrostatic force will be
[tex]F'=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F[/tex]
So the force has increased by a factor 4. By using [tex]F=8\cdot 10^{-9} N[/tex], we find
[tex]F'=4(8\cdot 10^{-9} N)=3.2\cdot 10^{-8} N[/tex]
The force of attraction between the soheres when they are 10cm apart is 3.2 * 10^-8 N
This is the force pulling bodies together given by
the force of attration, F = ( G q1 q2 ) / d^2
G refera to constant of attraction
q1 and q2 is the charges of the objects
d is the distance between the objects
G q1 q2 are all constant with respect to this question
let K = G * q1 * q2
F = k / d^2
for d = 20 cm = 0.2 m
F20 = k / 0.2^2
8*10^-9 = k / 0.2^2
K = 8*10^-9 *0.2^2
K = 3.2*10^-10
for d = 10 cm = 0.1 m
F10 = K / d^2
= (3.2*10^-10) * 0.1^2
= 3.2*10^-8 N
Read more on force of attraction here:
https://brainly.com/question/3150088
