Question

Two charged bodies exert a force of 0.245 N on each other. If they are moved so that they are one-fourth as far apart, what force is exerted

Answer 1

To solve this problem we will apply Coulomb's law. For which it tells us that if the charges are constant, the relationship between two forces is equivalent to the inverse of its squared distance. Mathematically this expression can be derived from the Coulomb equation for force:

$F = \frac{kq^2}{r^2}$

Here,

k = Coulomb's constant

q = Charge

r = Distance

$\frac{F_1}{F_2} = \frac{\frac{kq_1^2}{r_1^2}}{\frac{kq_2^2}{r_2^2}}$

$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$

$\frac{F_1}{F_2} = (\frac{r_2}{r_1})^2$

Under the considerations of the statement we have to

$F_1 = 0.245N$

$r_2= \frac{r_1}{4}$

Now replacing,

$\frac{0.245}{F_2} = (\frac{r_1/4}{r_1})^2$

$F_2 = \frac{0.245}{(1/4)^2}$

$F_2 = 3.92N$