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Why can even a small injury to the cornea have a major effect on vision?

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New force, $F'=48\times 10^3\ N$

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It is given that,

Force acting between two charged particles, $F=7.5\times 10^2\ N$

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

$F=k\dfrac{q_1q_2}{r^2}$ ............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

$F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}$ ..........(2)

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