Answer:
option (E)
Explanation:
F = 1 N
r = 2 cm
r' = 8 cm
F' = ?
According to the Coulomb's law
[tex]F \alpha \frac{1}{r^{2}}[/tex]
So, [tex]\frac{F'}{F}=\left (\frac{r}{r'} \right )^{2}[/tex]
[tex]\frac{F'}{1}=\left (\frac{2}{8} \right )^{2}[/tex]
F' = 0.063 N
When they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.
Given:
The point charges are initially at a distance of [tex]d=2[/tex] cm.
The initial force experienced by them is [tex]f=1[/tex] N.
The electrostatic force experienced by two charges q and Q is defined as,
[tex]f=K\dfrac{qQ}{r^2}[/tex]
where K is the constant and r is the distance between the charges.
The initial force experienced by them will be,
[tex]f=K\dfrac{qQ}{d^2}=1\rm\; N[/tex]
Now, the final distance between the charges is changed to 8 cm which is equal to 4d.
So, the new force on the charges will be,
[tex]F=K\dfrac{qQ}{(4d)^2}\\F=K\dfrac{qQ}{d^2}\times \dfrac{1}{16}\\F=f\times \dfrac{1}{16}\\F=0.0625\approx0.063\rm\;N[/tex]
Therefore, when they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.
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https://brainly.com/question/9774180
